Math 55 Ch 6 and 7 HW Qs

6.1 35d. How many one-to-one functions are there from a set with five elements to a set with a) 2 b) 5 c) 6 d) 7 elements

A function is one-to-one, if each domain element has a unique image (range). a) not possible d) 7⋅6⋅5⋅4⋅3=2520​

6.3 36. Suppose that a department contains 10 men and 15 women. How many ways are there to form a committee with six members if it must have more women than men?

4,5, or 6 women so C(15,4)C(10,2) + C(15,5)C(10,1) + C(15,6)C(10,0)

6.5 14. How many solutions are there to the equation x1 + x2 + x3 + x4 = 17, where x1, x2, x3, and x4 are nonnegative integers?

By theorem 2, the answer is C(4+17-1, 17) = C(20, 17)

6.5 32. How many different strings can be made from the letters in MISSISSIPPI, using all the letters?

By theorem 3 the answer is 11!/(4!4!2!)

Ch 7.2 12. Suppose that E and F are events such that p(E) = 0.8 and p(F) = 0.6. Show that p(E ∪ F) ≥ 0.8 and p(E ∩ F) ≥ 0.4.

First we have to prove P(E ∪ F) ≥ 0.8 which we can do using the definition ( A + B - AB) We apply theorem 2 from section 7.1. P(E∪F) must be <= to 1. This can be rewritten as: P(E) + P(F) - P(E ^ F) <= 1

Ch 7.1 36. Which is more likely: rolling a total of 8 when two dice are rolled or rolling a total of 8 when three dice are rolled?

Literally enumerate all possibilities and do the math.

MT 2 S16 Practice Exam How many poker hands (of five cards) contain exactly one ace, no king, and at least one heart?

Number of poker hands that contain one ace, no king = Number of poker hands that contain one ace, no king, AND 0 hearts + Number of poker hands that contain one ace, no king, AND 1,2,3,or 4 hearts. Take the converse to solve for case of at least one heart Total poker hands containing one ace, 4 non-king: 4C1 * 44C4 poker hands containing one ace, no king, and neither the ace nor any other of the 4 cards are kings or hearts: 1 non-heart ace: 3C1 0 non-heart king: 3C0 (is = to 4C0) = 1 4 non-heart, non-ace, non-king cards = 52-13 heart- 3 additional ace - 3 additional king

Ch 7.2 24. What is the conditional probability that exactly four heads appear when a fair coin is flipped five times, given that the first flip came up tails?

There are 16 equally likely outcomes of flipping a fair coin five times in which the first flip comes up tails (each of the other flips must come up either heads or tails so 2^4). Of these, only one results in four heads, namely THHHH. therefore the probability is 1/16

Ch 7.1 14. What is the probability that a five-card poker hand contains cards of five different kinds?

There are C(13,5) ways to choose the five kinds. For each card, there are 4 ways to choose it. Therefore there are C(13,5)*4^5 ways to choose the hand. The probability is that over 52C5

6.5 20.. How many solutions are there to the inequality x1 + x2 + x3 ≤ 11, where x1, x2, and x3 are nonnegative integers? [Hint: Introduce an auxiliary variable x4 such that x1 + x2 + x3 + x4 = 11.]

We introduce the non-negative variable x4, and our problem becomes the same as the problem of counting the number of nonnegative integer solutions to x1+x2+x3+x4 =11. By theorem 2, this is = to C(4+11-1,11)

Ch 7.1 44. Suppose that instead of three doors, there are four doors in the Monty Hall puzzle. What is the probability that you win by not changing once the host, who knows what is behind each door, opens a losing door and gives you the chance to change doors? What is the probability that you win by changing the door you select to one of the two remaining doors among the three that you did not select?

You had a 1/4 chance of winning with your original selection. Just as in the original problem, the host's actions did not change this, since he would act the same way regardless of whether your selection was a winner or a loser. Therefore, you have a 1/4 chance of winning if you do not change. This implies that there is a 3/4 chance of the prize being behind one of the other doors. Since there are 2 such doors, the chance of the prize being behind one of the doors is a total of 3/4, with there being an equal chance of it being behind each door (each door's probability is half of 3/4) which is 3/8 for each door. Therefore the answer is 3/8

6.1 48. In how many ways can a photographer at a wedding arrange 6 people in a row from a group of 10 people, where the bride and the groom are among these 10 people, if a) the bride must be in the picture? b) both the bride and groom must be in the picture? c) exactly one of the bride and the groom is in the picture?

a) 6* P(9,5) b) 6*5*P(8,4) c) 6*P(8,5) + 6*P(8,5) = 2(6*P(8,5)) = 80,640

6.5 46. In how many ways can a dozen books be placed on four distinguishable shelves a) if the books are indistinguishable copies of the same title? b) if no two books are the same, and the positions of the books on the shelves matter? [Hint: Break this into 12 tasks, placing each book separately. Start with the sequence 1, 2, 3, 4 to represent the shelves. Represent the books by bi , i = 1, 2, ... , 12. Place b1 to the right of one of the terms in 1, 2, 3, 4. Then successively place b2, b3,... , and b12.]

a) All that matters is the number of books on each shelf, so the answer is the number of solutions to x1+ x2 + x3+ x4 =12, where x_i is being viewed as the number of books on shelf i. The answer is therefore C(1+12-1,12) = 455 b) No generality is lost if we number the books b1, b2, ... b12 and think of placing book b1, then b2, and so on. There are clearly 4 ways to place b1, since we can put it as the first book on any shelf. After b1 is placed, there are 5 ways to place b2, since it can go to the right of b1, or it can be the first book on any of the shelves. There are 6 ways to place b3 (to the right of b1, to the right of b2, or as the first book on any of the shelves), 7 ways to place b4 (to the right of any of the three, or as the first on any of the shelves) 15 ways to place b12. Therefore the answer is the product of these numbers: 4*5*...*15 = 217,945,728,000

6.5: 10acd A croissant shop has plain croissants, cherry croissants, chocolate croissants, almond croissants, apple croissants, and broccoli croissants. How many ways are there to choose a) a dozen croissants? b) three dozen croissants? c) two dozen croissants with at least two of each kind? d) two dozen croissants with no more than two broccoli croissants? e) two dozen croissants with at least five chocolate croissants and at least three almond croissants?

a) C(6+12-1,12) = C(17,12) b) C(6+36-1,36) c) Two of each kind so 2*6 kinds = 12 croissants already picked. The remaining 12 can be picked in C(6+12-1,12) ways, which is the overall answer. d) Choosing at least 3 broccoli croissants: C(6+21-1,21), since once we've picked the three broccoli, there are 21 left to pick. Since there are C(6+24-1,24) ways to pick 24 croissants without restriction, there are C(6+24-1,24) - C(6+21-1,21) ways to choose no more than 2 broccoli. Alternatively u can just solve for number of ways with 0, 1, or 2 broccoli and sum that shit up. = 52,975

Ch 7.2 8. What is the probability of these events when we randomly select a permutation of {1, 2, ... , n} where n ≥ 4? a) 1 precedes 2. b) 2 precedes 1. c) 1 immediately precedes 2. d) n precedes 1 and n −1 precedes 2. e) n precedes 1 and n precedes 2

a) P(1 precedes 2) =1/2 b) by symmetry, also 1/2 c) we think of 1 and 2 being an object that is glued together, and thus, we are really permuting n-1 object -- the single numbers 3 to n, and the glued object 12/ There are (n-1)! to do this. Since there are n! permutations in all, the probability of randomly selecting one of these (n-1)! ways is (n-1)!/n! = 1/n

MT 2 2019 Practice 3a. How many ways are there to distribute six red hats, six blue hats and six gold hats to a group of 18 students if each student receives one hat? [The hats are identical except for their colors.] How many ways are there to carry out the task in part (a) if the two identical twins in the class are to receive the same color hat?

a. 18C6 * 12C6 * 6C6 b. 3 * 16C4 * 12C6 * 6C6

Ch 7.1 18. What is the probability that a five-card poker hand contains a straight flush, that is, five cards of the same suit of consecutive kinds?

apparently 40/C(52,5) because there are only 10*4 straight flushes (10 because ace up to 10 and 4 because thats the number of suits)

Ch 7.1 12. What is the probability that a five-card poker hand contains exactly one ace?

hypergeometric. G=4, g=1 B=48, b=4 N=52, n=5 abswer is 4*C(48,4) over C(52,5) = 0.3

MT 2 Prac 2: 3. There are 5 distinguishable bins labeled {1, 2, 3, 4, 5}. How many ways are there of placing 100 indistinguishable balls into the bins, if each bin must have at least as many balls as its label? Be sure to explain your reasoning

n indistinguishable objects into k distinguishable boxes: =C(n+k-1, k-1) = C(85+5-1, 5-1) = C(89,4)

6.4 5. How many terms are there in the expansion of (x + y)^100 after like terms are collected?

n+1, which is 101

6.1 40. How many subsets of a set with 100 elements have more than one element?

total number of subsets: 2^100 (corollary 6.4.1 on cheatsheet) total subsets with 0 elements: 1 total subsets with 1 element: 100 answer is therefore 2^100 - 101