STA 4.4 Counting
Combinations
Arrangements in which different sequences of the same items are counted as being the same (not separately). Example: The letter arrangements of abc, acb, bac, cab are all considered to be the same combination "Combinations Committee" reminds us that with members of a committee, rearrangements of the same members result in the same committee so the order does not count. If there are n different items available, and r of the n items are selected without replacement, and rearrangements of the same items are considered to be the same, then the number of combinations is given by the function nCr
Permutations
Arrangements in which different sequences of the same items are counted separately Example: arrangements of abc, acb, bac, cab are all counted separately as different permutations "Permutations Position" reminds us that with permutations, the positions of the items makes the difference
Multiplication counting rule
For a sequence of events in which the first event can occur n1, the second event can occur n2 ways, the third event can occur n3 ways, and so on, the total number of outcomes is n1 x n2 x n3 . . . . .
Fundamental counting rule
For a sequence of two events in which the first event can occur m ways and the second event can occur n ways, the events together can occur a total of m * n ways. Example Alarm system has a four digit code. The digits are 0 - 9. How many codes are possible? There can be 10 possible options for each digit of the code. 10 x 10 x 10 x 10 = 10,000 total possible options
Permutations rule (when some items are the same)
If there are n items with n1 alike, n2 alike,..., nk alike, the number of permutations of all n items is n! / n1!n2!...nk! Example 10 questions with 2 that are identical and 3 others that are also identical. We want the number of permutations 10! / 2!3! = 3,628,700 / 2x6 = 302,400 example: A classic counting problem is to determine the number of different ways that the letters of "etiquette" can be arranged. Find that number. When some items are identical to others, the permutations rule states that the number of different permutations (order counts) when n items are available and all n are selected without replacement, but some of the items are identical to others: n1 are alike, n2 are alike, and nk are alike, is given by the expression below. The factorial symbol (!) denotes the product of decreasing positive whole numbers. For example, 4!=4•3•2•1=24. By special definition, 0!=1. n!n1!n2!•••nk! There are 9 letters in the word "etiquette." Since the word "etiquette" contains 9 letters with 3 e's and 3 t's, use the permutations rule (when some items are identical) to count the number of different ways that the letters can be arranged. n!n1!n2!•••nk!=9!/3!3! Evaluate the expression. 9!/3!3!=10,080 Therefore, the number of different ways that the letters of "etiquette" can be arranged is 10,080.
Factorial rule
Number of ways n distinct objects can be arranged in order is given by Example 6! = 6x5x4x3x2x1 In calculator 6 MATH PRB #4 ENTER
Examples
There are 60 runners in a race. How many ways can the runners finish first, second, and third? Calculator 60 nPr 3 =205320 There are 205320 different ways that the runners can finish first through third. (Type a whole number.) _______________________________________________________ When testing for current in a cable with six color-coded wires, the author used a meter to test three wires at a time. How many different tests are required for every possible pairing of three wires? Calculator 6 nCr 3 The number of tests required is 20. ________________________________________________________ With a short time remaining in the day, a delivery driver has time to make deliveries at 7 locations among the 8 locations remaining. How many different routes are possible? Calculator 8 nPr 7 =40320 There are 40320 possible different routes. (Simplify your answer.) _____________________________________________________________ A corporation must appoint a president, chief executive officer (CEO), chief operating officer (COO), and chief financial officer (CFO). It must also appoint a planning committee with four different members. There are 17 qualified candidates, and officers can also serve on the committee. Complete parts (a) through (c) below. a. How many different ways can the officers be appointed? 17 nPr 4 = 57120 There are 57120 different ways to appoint the officers. b. How many different ways can the committee be appointed? 17 nCr 4 = 2380 There are 2380 different ways to appoint the committee. c. What is the probability of randomly selecting the committee members and getting the four youngest of the qualified candidates? P(getting the four youngest of the qualified candidates)= 1/2380 (Type an integer or a simplified fraction.) _____________________________________________________ How many different ways can the letters of "referred" be arranged? 8 letters total 3 r's repeated, 3 e's repeated 8!/3!3! 8! = 40320 3! = 6 40320/6*6 40320/36 =1120 The number of different ways that the letters of "referred" can be arranged is 1120. (Simplify your answer.) ______________________________________________________ Which of the following is NOT a requirement of the Combinations Rule, nCr=n!r!(n−r)!, for items that are all different? B. That order is taken into account (consider rearrangements of the same items to be different sequences). Your answer is correct.
Permutations Rule
When n different items are available and r of them are selected without replacement, the number of different permutations (order counts) is given by nPr. Rearrangements of the same items are counted as different permutations A permutation is an ordered arrangement of r objects chosen from n distinct objects without repetition nPr = n! / (n-r)! Exmaple Lottery probability is 1 / 5,245,786. This is when the numbers can be in any order. What is the probability of winning if the rules are changed that in addition to selecting the correct six numbers from 1 to 42, you must now select them in the same order as they are drawn? 42P6 = 42! / (42-6)! = 42! / 36! = 3,776,965,920 = P(winning)
Combinations rule
When n different items are available, but only r of them are selected without replacement, the number of different combinations (order does not matter) is found by nCr. nCr = n! / (n-r)!r! MATH -> PRB -> nCr Example Reduced staff from 32 to 28. 4 employees randomly selected. the 4 employees chosen are the oldest. Find the probability that when 4 employees are randomly selected from 32, that the 4 oldest are selected. Is it low enough to be reasonable? calculator nCr 32 nCr 4 P(4 oldest) = 1/32C4 =1/35960
