Probability

Experiment

A procedure that produces one outcome from a set of possible outcomes.

Poker Probability

A standard deck of 52 cards is divided into four suits (spades, clubs, hearts, diamonds), each with 13 kinds/ ranks(2 through 10, Jack, Queen, King, Ace).

Event (E)

A subset of the sample space, which means it's a collection of possible outcomes from the sample space that share a specific characteristic.

Among any group of 367 people, there must be at least two with the same birthday, because there are only 366 possible birthdays.

Among any group of 367 people, at least two have the same birthday because there are only 366 possible birthdays (including February 29). This is a straightforward application indicating that the number of "pigeons" (people) exceeds the number of "pigeonholes" (days in a year).

Permutation

An ordered arrangement of distinct objects from a set.

r-Permutation

An ordered arrangement of r elements selected from a set of n distinct elements, where the order of selection matters.

A student can choose a computer project from one of three lists. The three lists contain 23, 15, and 19 possible projects, respectively. No project is on more than one list. How many possible projects are there to choose from?

Application of Sum Rule: The total number of unique projects available across all lists is 23+15+19=57, as projects are distinct and not duplicated across lists.

How many ways are there to select five bills from a cash box containing $1 bills, $2 bills, $5 bills, $10 bills, $20 bills, $50 bills, and $100 bills? Assume that the order in which the bills are chosen does not matter, that the bills of each denomination are indistinguishable, and that there are at least five bills of each type.

Bills as Stars: Each selected bill is represented as a star (*). Types of Bills as Dividers: The different bill denominations are separated by dividers (|), representing the boundaries between different types of bills in the cash box. Given there are seven types of bills, there are six dividers needed to separate these types within the conceptual model. Selecting five bills translates into placing five stars among these dividers. so C(11,5)

Permutations with Repetition

Concept applies to situations where the same element can be chosen multiple times. If you're creating strings of length r from the 26 uppercase English letters, each position in the string can be any of the 26 letters, leading to 26^r possible strings. The number of r-permutations from a set of n objects with repetition allowed is n^r. his is derived using the product rule, where each of the r positions in the permutation can be filled in n ways.

Counting Variable Names in BASIC You need to calculate the number of valid variable names in a version of BASIC where names are 1 or 2 alphanumeric characters long, must start with a letter, and cannot be one of five reserved two-character strings.

Break Down Into Smaller Problems: Separate the problem into counting one-character names (V1) and two-character names (V2). Apply the Product Rule for Two-Character Names: First character: 26 options (must be a letter). Second character: 36 options (can be a letter or a digit).Total before exclusions: 26×36 Exclude Reserved Names: Subtract the 5 reserved names from the total for V2. Use the Sum Rule: Add V1 and V2 to get the total number of variable names. Step-by-Step Calculation V1: 26 (since there are 26 letters). V2: 26×36−5=931. Total: 26+931=957

Poker Hands from a Standard Deck: The number of 5-card poker hands from a standard 52-card deck is computed using combinations since the order of the cards does not matter. The calculation:

C(52,5) = 52!/5!*47!

EXAMPLE 9 How many ways are there to place 10 indistinguishable balls into eight distinguishable bins?

C(8 + 10 − 1, 10) = C(17, 10) --> 17!/10!7!

Combinations: Combinations deal with the selection of objects where the order does not matter. The formula for combinations, or r-combinations from a set of n distinct elements, is:

C(n,r) = n!/(r!(n-r)!) This formula is used when we are interested in the number of ways to select r elements from n without regard to the order of selection.

A sequence of 10 bits is randomly generated. What is the probability that at least one of these bits is 0?

Calculate complement because it would take a lot of math if we did for every event. Let E be the event that at least one of the 10 bits is 0. The Complement of E is the event that all the 10 bits are 1. Calculate the probability of complement E is straightforward. Every bit has a 1/2 chance of being 1 so (1/2)^10. --> 1/1024 Using Probability of Complements 1-1/1024 = 1023/1024

A group of 30 people have been trained as astronauts to go on the first mission to Mars. How many ways are there to select a crew of six people to go on this mission (assuming that all crew members have the same job)?

Choosing a 6-person crew from 30 candidates results in C(30,6)=593,775 possible selections

Assume that in a group of six people, each pair of individuals consists of two friends or two enemies. Show that there are either three mutual friends or three mutual enemies in the group.

Claim: The claim is that within this group, there must be either three people who are all friends with each other, or three people who are all enemies of each other. Simplified Explanation: Think of it like this: Choose any person from the group; let's call this person A. A knows five other people in the group, which creates five connections (like strings from A to each of the others). Now, these five connections can only be two types: a "friend" connection or an "enemy" connection. Imagine A has a magic hat with two compartments inside it: one for friends and one for enemies. A starts sorting the five connections into these compartments. According to the pigeonhole principle (which is like saying you can't fit more than a certain number of pigeons into a limited number of pigeonholes without some sharing a hole), one compartment must have at least three connections. Why? Because if you try to fit five connections into two compartments, one must end up with at least three (since 5 divided by 2 is 2.5, and since we can't have half a connection, we round up). If A has at least three friends (three connections in the "friend" compartment), we focus on them: let's call them B, C, and D. If any two of B, C, and D are friends, then together with A, they make a trio of friends. If none of B, C, and D are friends with each other, then they have to be enemies. Since they can't be A's enemies (they're in the friend compartment), they must be enemies with each other, making them a trio of enemies. The same logic applies if A has at least three enemies; we just look at those connections instead. So no matter how you slice it, A's connections lead to finding either a trio of friends or a trio of enemies in the group.

Telephone Numbering Plan. How many different North American telephone numbers are possible under the old plan and under the new plan?

Context: The North American numbering plan specifies formats for telephone numbers, with variations between the old plan (NYX-NNX-XXXX) and the new plan (NXX-NXX-XXXX). Old Plan: N = Any digit from value 2 through 9. so 8 Y = 0 or 1, so 2 X = 0 through 9, so 10 8*2*10 = 160 Office Code(NNX) = 8*8*10 = 640 Station Code(XXXX) = 10^4 = 10000 160*640*10000 + Total

Counting Bit Strings: Calculate the number of 8-bit strings that either start with a 1 or end with 00.

Count All Starting with 1: There are 2^7=128 ways, since the first bit is fixed and the remaining 7 can be anything. Count All Ending in 00: There are 2^6=64 ways, with the last two bits fixed and the first 6 can be anything. Count Overlap (Both Start with 1 and End with 00): There are 2^5=32 such strings, fixing both the first and last two bits. Apply Subtraction Rule: Subtract the overlap from the total to avoid double-counting. Total = 128+64-32 = 160

Majoring in Computer Science or Business: Given 350 applicants, with 220 majoring in computer science, 147 in business, and 51 in both, find how many majored in neither field

Count All in Computer Science or Business: Sum of those in either major. Subtract Overlap: Those majoring in both are counted in both the 220 and the 147. Find Neither: Subtract the total found from the overall number of applicants. Total neither=350−(220+147−51)=350−316=34

Seating Around a Circular Table Problem Statement: Find the number of distinct seating arrangements for four people around a circular table, considering arrangements identical if everyone has the same neighbors.

Count Linear Arrangements: Treat one seat as fixed to break the circle and arrange the remaining seats linearly, resulting in 4!=24 ways. Correct for Circular Symmetry: Since rotating the arrangement doesn't change the neighbors, each arrangement is counted 4 times (once for each person being in the 'first' seat). Apply Division Rule: Divide the total linear arrangements by the number of times each circular arrangement is counted. Distinct Arrangements = 4!/4=24/4=6

Suppose that there are 9 faculty members in the mathematics department and 11 in the computer science department. How many ways are there to select a committee to develop a discrete mathematics course at a school if the committee is to consist of three faculty members from the mathematics department and four from the computer science department?

Creating a committee with members from two different departments involves the product rule in combination with binomial coefficients, yielding C(9,3)⋅C(11,4)=27,720 ways to form the committee.

What is the probability that a positive integer selected at random from the set of positive integers not exceeding 100 is divisible by either 2 or 5?

Defining the Events: E1​: The number is divisible by 2. E2​: The number is divisible by 5. E1​∪E2​: The number is divisible by either 2 or 5. E1​∩E2​: The number is divisible by both 2 and 5, hence by 10. Counting the Outcomes: ∣E1∣: There are 50 numbers between 1 and 100 that are divisible by 2 (every other number). ∣E2∣: There are 20 numbers between 1 and 100 that are divisible by 5 (every fifth number). ∣E1∩E2∣: There are 10 numbers between 1 and 100 that are divisible by 10 (every tenth number, satisfying both divisibility criteria). Math: 50/100 = .5 20/100 = .2 10/100 = .1 p(E1​)+p(E2​)−p(E1​∩E2​)=0.5+0.2−0.1=0.6

How many ways are there to distribute hands of 5 cards to each of four players from the standard deck of 52 cards?

Distributing 5 cards to each of 4 players from a 52-card deck involves distinguishable objects (cards) going into distinguishable "boxes" (players' hands). 52!/(52-5-5-5-5)!5!5!5!5!

Show that among any n + 1 positive integers not exceeding 2n there must be an integer that divides one of the other integers.

Every number can be broken down into a bunch of twos multiplied by an odd number (like how 12=2^2 × 3). Now, because there are only n odd numbers less than 2n, but we have n+1 numbers in our list, two of them must have the same odd part (again, pigeonhole principle—more pigeons than holes). If you have two numbers with the same odd part, the one with more twos in it will divide the one with fewer twos (like 2^3 × 3 divides 2^2 × 3 because 2^3 is just 2^2 with an extra 2).

Suppose that a cookie shop has four different kinds of cookies. How many different ways can six cookies be chosen? Assume that only the type of cookie, and not the individual cookies or the order in which they are chosen, matters How many solutions does the equation x1 + x2 + x3 = 11 have, where x1, x2, and x3 are nonnegative integers

Example 4 (Cookie Shop): Demonstrates how to calculate the number of ways to choose six cookies from four types, leading to (9 6) or (9 3) ways, equaling 84. C(4 + 6 − 1, 6) = C(9, 3) Example 5 (Equation Solutions): Explains counting the solutions to x1 + x2 + x3 = 11 with nonnegative integers, equating it to (13 11) or (13 2), which is 78. It shows how counting C(3 + 11 − 1, 11) = C(13, 11) = C(13, 2)

Examples of Permutations

Example 4: Selecting a first, second, and third-prize winner from 100 people. There are P(100,3)=100×99×98=970,200 ways to do this. Example 5: Awarding gold, silver, and bronze medals to 8 runners. The number of ways is P(8,3)=8×7×6=336. Example 6: Planning a saleswoman's tour through 8 cities, starting from a specific city. There are 7!=5,040 ways for her to plan her tour through the remaining cities. Example 7: Arranging letters ABCDEFGH so that ABC always appears as a block. Treat ABC as a single object, leading to 6!=720 permutations.

Examples of Combinations

Example 8: Forming a committee of 3 students from a group of 4. There are C(4,3)=4 ways to form such a committee. Example 9: Identifying 3-combinations from the set {1,2,3,4}. There are C(4,3)=4 possible combinations: 1,2,3},{1,2,4},{1,3,4},{2,3,4}. Example 10: Finding 2-combinations of the set {a,b,c,d}. There are C(4,2)=6 combinations: {a,b},{a,c},{a,d},{b,c},{b,d},{c,d}.

Subsequences Within a Sequence. Any sequence of n^2+1 distinct real numbers contains a subsequence of length n+1 that is either strictly increasing or strictly decreasing.

Example Sequence: 8, 11, 9, 1, 4, 6, 12, 10, 5, 7. This sequence contains n^2+1=32+1=10 terms. According to the theorem, there must be a subsequence of length n+1=3+1=4 that is either strictly increasing or strictly decreasing. Strictly Increasing Subsequences of Length 4: 1, 4, 6, 12 1, 4, 6, 7 1, 4, 6, 10 1, 4, 5, 7 These are increasing because each subsequent number is larger than the previous one.

What probabilities should we assign to the outcomes H (heads) and T (tails) when a fair coin is flipped? What probabilities should be assigned to these outcomes when the coin is biased so that heads comes up twice as often as tails?

Fair coin --> p(H) = p(T) = 1/2 Biased Coin --> p(H) = 2p(T). Normalization Condition: p(H) + p(T) = 1 p(H) = 2p(T) we can substitute into the expression. 2p(T) +p(T) = 1 3p(T) = 1 p(T) = 1/3 Plug back into original relationship --> p(H) = 2p(T) --> p(H) = 2* 1/3 = 2/3

Find the probability that a hand of five cards in poker contains four cards of one kind?

Find the probability of getting a hand with four cards of the same rank (four of a kind) in poker. Step 1: Choose the rank for the four of a kind. There are 13 ranks (from Ace to King), so C(13,1)=13 ways to choose the rank. Step 2: Select all four cards of the chosen rank. There's only one way to do this since you need all four cards of the rank, so C(4,4)=1. Step 3: Choose the fifth card, which can be any of the remaining 48 cards (52 cards in the deck minus the 4 cards of the chosen rank), so C(48,1)=48. 13 × 1×48/C(52,5) = 0.00024

EXAMPLE 14 How many bit strings of length n contain exactly r 1s?

Finding the number of bit strings of length n with exactly r 1s translates to selecting r positions for the 1s out of n, which is C(n,r).

Bit Strings of Length Four Without Consecutive 1s Problem Statement: Count the number of 4-bit strings that do not contain consecutive 1s.

First Bit: Start with a root node that branches into two options: 0 or 1 (the first bit of the string). Subsequent Bits: For each branch, add two more branches to represent the options for the next bit, taking care to not add a branch for a 1 if the previous bit was also a 1. Count Leaves: The leaves of the tree (endpoints without further branches) represent all valid bit strings. Counting these gives the solution. The tree diagram would show 8 valid strings as its leaves, each following the rule against consecutive 1s.

Counting Subsets of a Finite Set Use the product rule to show that the number of different subsets of a finite set S is 2^|S|

For any finite set S, the number of its subsets can be represented as 2^∣S∣, where ∣S∣ is the cardinality (number of elements) of set S. Each element in S has two choices regarding its presence in a subset: it can either be included or excluded. Representing this choice with a bit string (where 1 means inclusion and 0 means exclusion), each element's choice corresponds to a bit in the string. For a set with n elements, the bit string has n bits. The number of possible bit strings of length n is 2^n, following the product rule (with 2 choices for each of n positions). Therefore, the total number of subsets of S is 2^∣S∣.

The Binomial Theorem

For any nonnegative integer n, and variables x and y, the expansion of (x+y)^n.

Stirling Numbers of the Second Kind

Formula: The number of ways to distribute n distinguishable objects into k indistinguishable boxes can be expressed as a sum of Stirling numbers of the second kind, covering all possible numbers of nonempty boxes from 1 to k: For specific n and j, Stirling numbers can be found using the inclusion-exclusion principle, represented by the formula. This formula encapsulates the combinatorial essence of distributing objects into boxes, factoring in the permutations of objects and the indistinguishability of the boxes through alternating sums.

Playoff Series Outcome Determine the number of ways a best-of-five playoff series can be won, assuming a team wins the series by winning three games.

Games as Stages: Each level of the tree represents a game in the series. Branches for Outcomes: From each node, branch out to two nodes for the next game, one for each team winning the current game. Terminate Early: Branches end as soon as one team wins three games, as the series concludes. Count Paths to Leaves: The paths from the root to leaves represent possible series outcomes. Counting these paths gives the total number of ways the series can play out. For this scenario, the tree diagram reveals 20 distinct paths or ways the playoff can occur.

DNA and RNA molecules encode genetic information through sequences of nucleotides, where the arrangement and combination of these nucleotides (with bases A, T, C, G for DNA and A, U, C, G for RNA) determine the structure of proteins.

Genetic Encoding: The sequence of bases (A, C, G, T for DNA; A, C, G, U for RNA) encodes genetic information. In DNA, sequences of three bases (codons) are required to specify one of the 22 essential amino acids needed to build proteins. There are 64 possible codons (4^3 = 64), which is sufficient to encode all 22 amino acids. Genome and Genes: A genome is the complete set of genetic material of an organism, including all of its genes. A gene is a segment of DNA that encodes for a specific protein. Genes are the functional units of heredity. Size of Genomes: Simpler organisms like algae and bacteria may have between 100,000 (10^5) and 10,000,000 (10^7) bases in their DNA. More complex organisms like insects, birds, and mammals can have between 100,000,000 (10^8) and 10,000,000,000 (10^10) bases. This demonstrates the vast complexity and variability of genetic information across species.

Probability of the Union of Two Events

How to find the probability that at least one of two events occurs in a given sample space. p(E1 U E2) = p(E1) +p(E2) - p(E1n E2)

Counting IPv4 Internet Addresses. Calculate the number of available IPv4 addresses, considering the different class restrictions and exclusions.

IPv4 Address Calculation Understanding IPv4 Addresses An IPv4 address is 32 bits long, consisting of a network identifier (netid) and a host identifier (hosted). Address Classes - Class A: Begins with a '0', followed by a 7-bit netid and a 24-bit hostid.- Class B: Begins with '10', followed by a 14-bit netid and a 16-bit hostid.- Class C: Begins with '110', followed by a 21-bit netid and an 8-bit hostid. Special Cases and Restrictions Certain netids and hostids are restricted or reserved for special uses, such as the netid '1111111' which is not available for Class A, and hostids that are all zeros or all ones are not usable. Calculating Available Addresses - Class A Addresses: 127 usable netids (128 total minus 1 special case) each having 16,777,214 usable hostids (16,777,216 total minus 2 special cases). Total Class A Addresses: 127 netids × 16,777,214 hostids = 2,130,706,178 addresses. - Class B Addresses: 16,384 netids each having 65,534 usable hostids (65,536 total minus 2 special cases). Total Class B Addresses: 16,384 netids × 65,534 hostids = 1,073,709,056 addresses. - Class C Addresses: 2,097,152 netids each having 254 usable hostids (256 total minus 2 special cases). Total Class C Addresses: 2,097,152 netids × 254 hostids = 532,676,608 addresses. Total Available IPv4 Addresses The total sum of available addresses across Class A, B, and C is 2,130,706,178 (Class A) + 1,073,709,056 (Class B) + 532,676,608 (Class C) = 3,737,091,842 available IPv4 addresses.

What is the probability that when two dice are rolled, the sum of the numbers on the two dice is 7?

Identify the Total Number of Possible Outcomes: Each die has 6 faces, so a single die has 6 possible outcomes. When two dice are rolled, the total number of possible outcomes is the product of the number of outcomes for each die: 6×6=36. This uses the principle of the product rule in combinatorics. Determine the Successful Outcomes: A successful outcome is one where the sum of the two dice equals 7. The pairs that satisfy this condition are: (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), and (6, 1). Each pair represents the outcome of the first die and the second die, respectively. Calculate the Probability: The probability of an event is the number of successful outcomes divided by the total number of possible outcomes. With 6 successful outcomes and 36 possible outcomes, the probability p is calculated as: 6/36

Generalized Pigeonhole Principle

If N objects are placed into k boxes, then at least one box contains at least ⌈n/k​⌉ objects, where ⌈x⌉ denotes the smallest integer greater than or equal to x. This principle can be demonstrated through contraposition: if it were not true that at least one box contains ⌈n/k⌉ objects, then the maximum number of objects that could be distributed without exceeding this number in any box would be less than N, which contradicts the premise that there are N objects.

Extended Product Rule

If a procedure involves performing m tasks in sequence and each task Ti (for i = 1,2,..m) can be done in ni ways then there are n1*n2*nm ways to complete the prodcure

Counting One-to-One Functions How many one-to-one functions are there from a set with m elements to one with n elements?

If m > n, no one-to-one functions exist. If m <= n , the number of one-to-one functions is calculated by reducing the choice for each subsequent element, resulting in n*(n-1)*...*(n-m+1) combinations.

Uniform Distribution

In a scenario where each outcome is equally likely, the probability of each outcome is 1/n, where n is the total number of outcomes in the sample space S.

In any group of 27 English words, there must be at least two that begin with the same letter, because there are 26 letters in the English alphabet.

In any group of 27 English words, at least two must start with the same letter, given that there are only 26 letters in the English alphabet. This demonstrates the principle in the context of alphabetic categorization.

k := 0 for i1 := 1 to n1 for i2 := 1 to n2 ... for im := 1 to nm k := k + 1

Initial Value: K is initialized to 0 Nested Loops: M nested loops. Each loop Ij iterates from 1 to NJ where j ranges from 1 to m Incrementing K: Inside innermost loop, k is incrementing by 1 with each iteration of the entire set of nested loops Example with Specific Numbers Suppose you're asked to find the value of k for n1​=3, n2​=2, and n3​=4. Here's how you would solve it: Identify each loop's range: The first loop iterates 3 times (n1​=3). The second loop iterates 2 times (n2​=2). The third loop iterates 4 times (n3​=4). Calculate the total number of iterations: The total iterations are 3×2×4=24. Determine the final value of k: Since k starts at 0 and is incremented once for each iteration of the nested loops, after all iterations are completed, k will be incremented 24 times. Thus, the final value of k is 24.

k := 0 for i1 := 1 to n1 k := k + 1 for i2 := 1 to n2 k := k + 1 ... for im := 1 to nm k := k + 1

K starts at 0 Each loop represents a separate block of code that increments k by 1 during each iteration. There are no nested loops, meaning each loop runs independently and sequentially after the previous loop completes. The sum rule is applied here because the loops are independent tasks that contribute to increasing k. Since each loop increments k a specific number of times (n i ​ for the i^th loop), the total increment of k across all loops is the sum of all increments. Calculating the Final Value of k to find the final value of k, sum the total number of increments from each loop: 1 n 1 ​ +n 2 ​ +...+n m ​. This sum gives us the total number of times k has been incremented across all loops. -------------------------------------------------- Example: Given: n1​=3, n2​=5, n3​=2 (for an example with m=3 loops) Solution: Calculate the total increments to k: 3+5+2=10. Thus, the final value of k after all loops have executed is 10.

k := 0 for i1 := 1 to n for i2 := 1 to i1 ... for im := 1 to im−1 k := k + 1

Loop Dependency: Notice that each loop's range is constrained by the previous loop's index, starting from 11 to n. This dependency ensures that the values of i1​,i2​,...,im​ form a non-decreasing sequence. Increment Operation: The operation k:=k+1 occurs once for every complete iteration of the innermost loop, indicating that k counts the total number of iterations across all nested loops. For example, if n=4 and m=3 (indicating three nested loops with the outermost ranging up to 4), you would calculate: C(4+3-1,3) = C(6,3) = 6!/3!3!

Subtraction rule

Method to accurately count the total number of unique ways a task can be completed when some options overlap between two different methods. This rule is essential for correcting overcounts that occur when the sum rule is naively applied to cases with common elements between the sets or options being considered.

Key Principles for Assigning Probabilities Let S be the sample space of an experiment with a finite or countable number of outcomes. We assign a probability p(s) to each outcome s.

Non-Negativity and Upper Bound: Every outcome s in the sample space S must have a probability p(s) that is between 0 and 1, inclusive. This requirement ensures that probabilities are meaningful and quantifiable. Normalization: The sum of the probabilities of all possible outcomes must equal 1. This condition reflects the certainty that one of the outcomes in the sample space will occur when the experiment is performed. The set of probabilities assigned to all possible outcomes of an experiment forms a probability distribution. This distribution must adhere to the above principles, whether the sample space is finite, countable, or infinite.

Calculate the probability of drawing the numbers 11, 4, 17, 39, and 23 in that order from a bin of 50 balls.

Solution (a) Without Replacement: The sequence of drawing reduces the total available choices by one each time (50, then 49, 48, 47, 46). Total ways: 50×49×48×47×46. 1/50×49×48×47×46. Solution (b) With Replacement: Each draw is independent with 50 choices each time, even after drawing a ball. 1/50^5

Suppose that a die is biased (or loaded) so that 3 appears twice as often as each other number but that the other five outcomes are equally likely. What is the probability that an odd number appears when we roll this die?

Number 3 is twice as likely to appear compared to the other numbers (1,2,4,5,6) p(1)=p(2)=p(4)=p(5)=p(6)=1/7 p(3) = 2/7 (7 because 6 sided die + the extra probability of 3) 1/7+2/7+1/7 = 4/7

Calculating 2-permutations of set S={a,b,c} gives 6 different arrangements,

P(3,2)=3⋅2=6.

Selecting 3 students out of 5 for a picture in line (order matters): Arranging all 5 students in a line

P(5,3)=5⋅4⋅3=60 P(5,5)=5!=120

Permutations

Permutations Permutations focus on the ordered arrangement of objects. For selecting r elements out of a set of n distinct elements, the formula for calculating permutations is: P(n,r)=n!/(n−r)! This formula essentially states that if you want to arrange r objects from a set of n, you start with n options for the first object, n−1 for the second, and so on, until you have selected r objects.

For set S={1,2,3}

Permutations include 3, 1, 2, and a 2-permutation example is 3, 2.

Sum rule

Rule states that if a task can be done in one of n 1 ​ways or in one of n 2 ​ ways, and these two sets of ways do not overlap (i.e., they are mutually exclusive), then the task can be done in n 1 ​ +n 2 ​ ways in total.

An urn contains four blue balls and five red balls. What is the probability that a ball chosen at random from the urn is blue?

Sample Space (S): All the balls in the urn, totaling nine (four blue and five red). Event (E): Drawing a blue ball from the urn. Number of Outcomes in E: Four, since there are four blue balls. Total Number of Possible Outcomes in S: Nine (the total number of balls). P(E) = 4/9

Selecting 47 Cards

Selecting 47 cards from a 52-card deck is equivalent to choosing 5 to leave out, so (52,47)=(52,5), which also equals 2,598,960. This symmetry illustrates a fundamental property of combinations, where selecting r items from n is the same as leaving out n−r items.

Example 12 (Tennis Team Selection): How many ways are there to select five players from a 10-member tennis team to make a trip to a match at another school?

Selecting 5 players from a 10-member tennis team results in C(10,5)=252 ways. This scenario showcases the direct application of the combination formula for selecting a subset where the order doesn't matter.

Suppose that either a member of the mathematics faculty or a student who is a mathematics major is chosen as a representative to a university committee. How many different choices are there for this representative if there are 37 members of the mathematics faculty and 83 mathematics majors and no one is both a faculty member and a student?

Since a representative can be chosen from either group without any overlap, the total number of choices is the sum of the two options: 37+83=120 possible representatives.

"I Love New Jersey" T-Shirts Calculate the total number of different "I Love New Jersey" T-shirts a shop must stock, considering size and color variations with specific restrictions on XL and XXL sizes.

Sizes as First Decision Point: Begin with branches for each shirt size (S, M, L, XL, XXL). Colors as Subsequent Choices: For each size, branch out to represent available colors. Note the restrictions on XL and XXL sizes and adjust branches accordingly. Count Leaves: Each leaf represents a unique size-color combination. Counting the leaves gives the total inventory requirement. The tree for this example shows that the shop needs to stock 17 different T-shirts to cover all size and color combinations given the restrictions.

Small Prize Probability vs Large Prize Probability

Small With nine possible wrong digits for the mismatched position and only one possible combination for each of the correctly matched positions, there are 99 ways to achieve a match for each specific position of the mismatch. Since there are four possible positions for the mismatch, the total number of ways to win the small prize is 4×9=36, leading to a probability of 36/10,000=9/2,500=0.0036. Large The total number of possible combinations is 10^4=10,000, given by the product rule (multiplying the number of outcomes at each stage). Since there is only one correct sequence, the probability of winning the large prize is 1/10,000=0.0001.

Counting Valid Passwords Find the number of valid passwords between 6 to 8 characters long, using uppercase letters and digits, with at least one digit required.

Solution Strategy Calculate Total Options Without Restrictions: Use the product rule for total strings of lengths 6, 7, and 8. Subtract Invalid Options: Compute strings without digits (only letters) and subtract from the total for each length. Combine Results Using the Sum Rule: Add the valid counts for lengths 6, 7, and 8. Step-by-Step Calculation For 6 Characters: Total = 36^6, Invalid =26^6, Valid =36^6−26^6. Repeat for 7 and 8 Characters: Apply the same formula. Total Valid Passwords: Sum of valid passwords of lengths 6, 7, and 8.

INDISTINGUISHABLE OBJECTS AND INDISTINGUISHABLE BOXES Some counting problems can be solved by determining the number of ways to distribute indistinguishable objects into indistinguishable boxes. How many ways are there to pack six copies of the same book into four identical boxes, where a box can contain as many as six books?

Solution:We will enumerate all ways to pack the books. For each way to pack the books, we will list the number of books in the box with the largest number of books, followed by the numbers of books in each box containing at least one book, in order of decreasing number of books in a box. The ways we can pack the books are 6 5, 1 4, 2 4, 1, 1 3, 3 3, 2, 1 3, 1, 1, 1 2, 2, 2 2, 2, 1, 1

(n+r-1, r)

Stars and Bars Representation: This method visualizes a problem of selecting r items from n categories with repetition allowed. Here, "stars" represent the items chosen, and "bars" (or dividers) represent the boundaries between different categories. Why n+r−1: In this formula, n−1 represents the number of dividers needed for n categories, and r is the number of items to be chosen. The total number of positions (stars and bars combined) is n−1+r, out of which you are choosing positions for either the r stars or the n−1 bars.

Find the probability of getting a full house in poker, which consists of three cards of one rank and two cards of another rank.

Step 1: Choose two ranks. Order matters because three of one kind and two of another are different outcomes. The number of ways is P(13,2)=13×12. Step 2: Choose three cards from the first rank C(4,3)=4) and two cards from the second rank C(4,2)=6). Divide by The total number of possible 5-card hands from a 52-card deck is C(52,5)

DISTINGUISHABLE OBJECTS AND INDISTINGUISHABLE BOXES Counting the ways to place n distinguishable objects into k indistinguishable boxes is more difficult than counting the ways to place objects, distinguishable or indistinguishable objects, into distinguishable boxes. How many ways are there to put four different employees into three indistinguishable offices, when each office can contain any number of employees?

The challenge here involves distributing four distinguishable employees (A, B, C, D) into three indistinguishable offices. The solution requires a detailed enumeration of possible distributions, considering the indistinguishable nature of the offices. All in One Office: Only 1 way (A,B,C,D}). Three in One Office, One in Another: 4 ways, since any of the four employees can be in the office by themselves. Two in One Office, Two in Another: 3 ways, based on pairing any two employees together. Two in One Office, One Each in the Others: 6 ways, reflecting all possible distributions of the remaining employees once a pair is chosen for one office. By counting all possibilities, we find 14 distinct ways to distribute the employees, demonstrating the need for enumeration in such problems due to the lack of a straightforward formula for cases involving indistinguishable containers. Stirling Numbers of the Second Kind To formalize and generalize the counting of distributions of distinguishable objects into indistinguishable boxes, we use Stirling numbers of the second kind, S(n,j). These numbers count the ways to partition a set of n distinguishable items into j nonempty subsets (or indistinguishable boxes).

Sample Space (S)

The complete set of all possible outcomes of an experiment.

Pigeonhole Principle Theorem: If k is a positive integer and k+1 or more objects are placed into k boxes, then at least one box contains two or more objects.

The contrapositive of the statement "If k+1 or more objects are placed into k boxes, then at least one box contains two or more objects" is "If none of the k boxes contains more than one object, then there are at most k objects." Assume that each of the k boxes contains at most one object. Under this assumption, the maximum number of objects that can be accommodated is k, one per box. Contradiction: There are k+1 objects, implying at least one box must contain more than one object, contradicting the assumption.

Division Rule

The division rule addresses scenarios where a task can be done in multiple ways, but each distinct way is counted multiple times across different methods. It corrects for this by dividing the overcounted total by the number of times each unique option is counted.

The Division Rule in Combinations

The division rule in combinations ensures that each combination is counted once by dividing the total number of permutations by the number of ways to arrange the r elements within each combination. Calculating Combinations C(n,r) = n(n-1)...(n-r+1)/r!

Combinatorial Proofs of Binomial Coefficient Identities

The identity C(n,r)=C(n,n−r) captures the essence of the symmetry in combinations. This can be proven using combinatorial logic: Bijective Proof: Every subset of S with r elements naturally corresponds to a complementary subset of n−r elements (and vice versa), implying a one-to-one correspondence between the sets of r-element subsets and n−r-element subsets of S. This bijection ensures that the number of r-combinations is equal to the number of n−r-combinations. Double Counting Proof: The total number of ways to choose r elements from a set S is the same as choosing which n−r elements to leave out, illustrating that both sides of the identity count the same quantity in two different ways.

Corollary 2: Combinatorial Symmetry

The identity C(n,r)=C(n,n−r) reflects the symmetric nature of combinations. It tells us that the number of ways to choose r items from n is the same as the number of ways to exclude r items from n, a principle that simplifies many counting problems.

Binomial Coefficients

The number of ways to choose r elements from a set of n elements, denoted as (n/r​) or "n choose r."

Probability of an Event E

The probability of an event E is defined as the sum of the probabilities of the outcomes that make up E. This is a key principle that holds regardless of whether the distribution is uniform or not.

Probability of Complements

The probability of the complement of an event E (denoted as ˉEˉ or E ′ ) is 1 minus the probability of E itself. Given a sample space S and an event E, the complement of E (ˉEˉ) includes all outcomes in S that are not in E. The probabilities of E and ˉEˉ together cover the entire sample space S, so p(E)+p(ˉEˉ)=1,

Probability of an Event (p(E))

The ratio of the number of outcomes in the event (|E|) to the total number of outcomes in the sample space (|S|), expressed as p(E) = |E| / |S|.

What is the least number of area codes needed to guarantee that the 25 million phones in a state can be assigned distinct 10-digit telephone numbers? (Assume that telephone numbers are of the form NXX-NXX-XXXX, where the first three digits form the area code, N represents a digit from 2 to 9 inclusive, and X represents any digit. Given 8 million unique numbers per area code.

There are eight million different phone numbers of the form NXX-XXXX |25,000,000/8000000| = 4

Combinations with Repetition

These are used when the order of selection doesn't matter, but elements can be chosen more than once. Selecting four pieces of fruit from a bowl that contains at least four pieces each of apples, oranges, and pears. The solution involves listing all possible selections where the type of fruit matters but not the specific pieces, leading to 15 combinations. To count r-combinations from an n-element set where repetition is allowed, a method involving partitioning or a formula is used. The concept is sometimes explained using "stars and bars" or "balls and urns" techniques, which provide a way to visualize and calculate the combinations.

Counting Indistinguishable Elements

This area addresses counting arrangements of items where some or all of the items are identical. Example: Counting the rearrangements of the word "SUCCESS" involves dealing with repeated letters. The solution involves dividing the total arrangements by the factorial of the number of each repeated element to account for indistinguishability.

Placing Distinguishable Elements in Boxes

This involves counting the number of ways distinct items can be distributed into containers, which may have restrictions like "at least one item per box" or "boxes are indistinguishable." Example: Dealing out poker hands to four players involves distributing a deck of distinct cards into four parts, one for each player.

Calculating Permutations

To calculate the number of ways to arrange n students from a group of n students in a line (which is an r-permutation of the group), we use the product rule, taking into account that each selection reduces the number of available choices by one. Formula for r-Permutations Given a set with n distinct elements, the number of r-permutations, denoted as P(n,r), can be calculated using the product rule as follow P(n,r) = n*(n-1)*(n-2)*....*(n-r+1) Simplifying the Formula n!=n*(n-1)*(n-2)*...*1 p(n,r)=n!/(n-r)!

What is the minimum number of students required in a discrete mathematics class to be sure that at least six will receive the same grade, if there are five possible grades, A, B, C, D, and F?

To ensure at least six students receive the same grade out of five possible grades, you need ⌈N/5​⌉=6 students, leading to N=5(6−1)+1=26 as the minimum.

How many students must be in a class to guarantee that at least two students receive the same score on the final exam, if the exam is graded on a scale from 0 to 100 points?

To ensure that at least two students receive the same score on a final exam graded from 0 to 100, there must be at least 102 students. This is because there are 101 possible scores (the pigeonholes), so with 102 students (the pigeons), at least one score must be assigned to two or more students.

a) How many cards must be selected from a standard deck of 52 cards to guarantee that at least three cards of the same suit are chosen? A standard deck of 52 cards has 13 kinds of cards, with four cards of each of kind, one in each of the four suits, hearts, diamonds, spades, and clubs. b) How many must be selected to guarantee that at least three hearts are selected?

To guarantee at least three cards of the same suit, you must select 9 cards. This follows since dividing any more than 8 cards among the 4 suits inevitably results in one suit having 3 cards. |N/4| <= 3 N=9 To ensure selecting at least three hearts, up to 42 cards might need to be drawn, considering the worst-case scenario where all other suits are drawn first. Total Hearts: There are 13 hearts in the deck. Total Non-Hearts: There are 39 non-heart cards (diamonds, clubs, spades). Must select 42 (39+3) to get 42 cards so 3

Monty Hall Problem

Understanding the Monty Hall Problem You're on a game show and given a choice of three doors: behind one door is a prize, and behind the other two, there's nothing. After you choose a door, the host, who knows what's behind the doors, opens another door that does not have the prize. At this point, you're given the choice to stick with your original selection or switch to the other unopened door. Initial Probabilities Choosing the Prize Initially: The probability of choosing the door with the prize on your first choice is 1/3, as there are three doors and only one has the prize. Choosing a Losing Door Initially: The probability that you initially chose a door without the prize is 2/3. After the Host Opens a Door The host's action of opening a door (which always reveals a door without the prize) is designed to confuse the contestant. However, the host's action doesn't change the initial probabilities. If you initially chose the winning door (1/3 chance), switching would make you lose. If you initially chose a losing door (2/3 chance), switching would make you win, because the host's action of opening a door has effectively removed the other losing option from consideration. Why Switching Increases Your Chances Without Switching: Your chance of winning stays at the initial 1/3, since you're sticking with your first choice. -With Switching: The crucial insight is that if you initially chose a losing door (2/3 chance), the host's intervention ensures that switching doors will now lead you to the prize. Therefore, switching doors gives you a 2/3 chance of winning.

Indistinguishable Objects and Boxes

When objects and boxes are indistinguishable, you enter the realm of partition problems and packing problems, which often lack simple closed formulas but are rich in combinatorial insights.

There are many lotteries now that award enormous prizes to people who correctly choose a set of six numbers out of the first n positive integers, where n is usually between 30 and 60. What is the probability that a person picks the correct six numbers out of 40?

When picking six numbers out of 40, the situation is different because the order in which the numbers are chosen does not matter. C(n,k)=n!/[k!(n−k)!], where n is the total number of items to choose from, k is the number of items to choose, and n! denotes the factorial of n. Thus, C(40,6)=3,838,380, and the probability of selecting the winning combination is 1/3,838,380≈0.00000026.

Product Rule

When procedure can be divided into a sequence of tasks, if there are n1 ways to do the first task and n2 ways to do the second task (ASSUMING FIRST TASK IS ALREADY DONE), then there are n1 * n1 ways to complete the procedure Office Assignment: A new company with just two employees, Sanchez and Patel, rents a floor of a building with 12 offices. How many ways are there to assign different offices to these two employees? For two employees and 12 offices, with each needing a different office, there are 12×11=132 ways to assign offices. Auditorium Chair Labeling: The chairs of an auditorium are to be labeled with an uppercase English letter followed by a positive integer not exceeding 100. What is the largest number of chairs that can be labeled differently? Labeling chairs with an uppercase letter and a number up to 100, results in 26×100=2600 unique labels. Microcomputer Ports: There are 32 microcomputers in a computer center. Each microcomputer has 24 ports. How many different ports to a microcomputer in the center are there: With 32 microcomputers, each having 24 ports, there are 32×24=768 different port selections available. Bit Strings: How many different bit strings of length seven are there? For bit strings of length seven, each bit having two possible values (0 or 1), there are 2^7=128 different bit strings.

Permutations with Indistinguishable Objects: The Principle

When you have a set of objects where some are indistinguishable from others (i.e., there are duplicates), the straightforward n! permutations formula overcounts the distinct arrangements. The correct approach divides out the overcounting for each set of indistinguishable objects. The word "SUCCESS" has 7 letters with repetitions: three S's, two C's, one U, and one E. To find distinct rearrangements: Start with Total Permutations: Without considering repetitions, there would be 7! ways to arrange 7 letters. Adjust for Repetitions: Divide by the factorial of the count of each repeated letter to correct for overcounting. For 3 S's, divide by 3! For 2 C's, divide by 2! U and E are unique, so no division is needed for them. 7!/3!2!1!1!=420

Among 100 people there are at least |100/12| = 9 who were born in the same month

With 100 people, there are at least ⌈100/12​⌉=9 who were born in the same month, considering there are 12 months (boxes) to distribute 100 people (objects).

A coin is flipped three times, with all eight outcomes equally likely. Given that the first flip comes up tails (F), we want to find the probability that an odd number of tails appears (E).

With the first flip being tails, the sample space is reduced to F = {TTT,TTH,THT,THH}. Since each outcome in F is equally likely --> 1/4 For an odd number of tails, we have E ∩ F = {TTT,THH} Since we only have two outcomes 1/4 + 1/4 = 2/4 = 1/2 is the answer.

Suppose that a computer science laboratory has 15 workstations and 10 servers. A cable can be used to directly connect a workstation to a server. For each server, only one direct connection to that server can be active at any time. We want to guarantee that at any time any set of 10 or fewer workstations can simultaneously access different servers via direct connections. Although we could do this by connecting every workstation directly to every server (using 150 connections), what is the minimum number of direct connections needed to achieve this goal?

Workstations: 15 total, labeled W1 through W15. Servers: 10 total, labeled S1 through S10. Requirement: Any 10 workstations must be able to connect simultaneously to 10 different servers. Proposed Solution Directly connect each of the first 10 workstations (W1 through W10) to a unique server (W1 to S1, W2 to S2, ..., W10 to S10). This ensures each of these workstations always has a unique server. For the remaining 5 workstations (W11 through W15), connect each of these to all 10 servers. This guarantees that these workstations can always access a server, even if the first 10 are all occupied. Total Connections Direct Connections for W1 to W10: 10 connections (each to a different server). Direct Connections for W11 to W15: Each connects to all 10 servers, yielding 5 workstations × 10 servers = 50 connections. Grand Total: 10 (from W1-W10) + 50 (from W11-W15) = 60 connections. Justification Unique Access for Any 10 Workstations: Any subset of 10 workstations can always access 10 unique servers. If the subset includes any workstations from W11 to W15, these can shift to any unused server because they are connected to all servers. Optimality Argument: If we have fewer than 60 connections, then at least one server connects to 5 or fewer workstations (since 59 connections divided evenly among 10 servers results in 5.9 connections per server, implying at least one server has 5 or fewer connections). If a server has only 5 connections, then there are only 9 other servers to handle possibly 10 workstations wanting simultaneous connections, which is impossible if all need separate servers.

During a month with 30 days, a baseball team plays at least one game a day, but no more than 45 games. Show that there must be a period of some number of consecutive days during which the team must play exactly 14 games.

Write down the cumulative number of games played after each day (1st day, 2nd day, etc.). Now, add 14 to each of these numbers (because we're looking for a stretch where 14 games are played). You'll have two lists of numbers: one is just the cumulative totals, and the other is each of those totals plus 14. Since there are 30 days, each list has 30 numbers. But, because they play no more than 45 games, all these numbers are between 1 and 59. You've got 60 numbers total, but only 59 possible outcomes (from 1 to 59). So, at least two numbers must be the same in these lists (this is where the pigeonhole principle comes in—it's like having 60 pigeons but only 59 holes for them to go into). The only way you can have the same number in both lists is if there's a stretch of days where exactly 14 games are played (because we made one list by adding 14 to the totals).

Theorem 3 generalizes the principle used in Example 7 for any number of objects and any number of indistinguishable sets within those objects. If you have n objects total, among which there are n1​ of one type, n2​ of another, ..., nk​ of the kth type, then the number of distinct permutations is:

n!/n1!*n2!..nk!

Counting Functions How many functions are there from a set with m elements to a set with n elements?

n^m

A bit string of length four is generated at random so that each of the 16 bit strings of length four is equally likely. What is the probability that it contains at least two consecutive 0s, given that its first bit is a 0? (We assume that 0 bits and 1 bits are equally likely.)

p(E|F) = p(E∩F)/p(F) E∩F = {0000, 0001, 0010, 0011, 0100} This gives 5/16 p(F) --> 2^3 = 8 possibilities as first bit is fixed so 8/16 --> 1/2 Or we have 8 bit string lengths that start with 0 5/16 / 1/2 --> 5/8

What is the conditional probability that a family with two children has two boys, given they have at least one boy? Assume that each of the possibilities BB, BG, GB, and GG is equally likely, where B represents a boy and G represents a girl. (Note that BG represents a family with an older boy and a younger girl while GB represents a family with an older girl and a younger boy.)

p(E|F) = p(E∩F)/p(F) {BB, BG, GB, GG} E = {BB} F = {BB, BG, GB} p(E∩F) --> Interaction {BB} = 1/4 p(F) --> 3/4 1/4 / 3/4 = 1/3