Homework 9 Solutions

Transcription

1 Homework 9 Solutions 6.2: 2: There are 26 letters each student s last name could start with, so by the Pigeonhole Principle there must be some letter that at least 30 2 students have beginning both 26 their last names. 6.2: : a She must take 5 balls: taking balls isn t enough she could get 2 of each and if she takes 5 the Pigeonhole Principle guarantees she will have at least 5 3 balls of the 2 same color. b She must take 3: in the worst case she takes all the red balls first. 6.2: 6: There are d possible remainders when divided by d namely the integers 0,, 2,... d. By the Pigeonhole Principle, some remainder must be hit twice if we have d + integers, so there exist at least two integers in our set with the same remainder. 6.2: 8: Here our holes will be values in T. Because S > T, the Pigeonhole Principle tells us that there must be at least 2 elements pigeons in S mapping to the same value hole in T. Thus there are s, s 2 S with fs fs : 0: The midpoint of the line between x i, y i and x j, y j is x i +x j, y i+y j 2 2. These coordinates will be integers as long as x i and x j have the same parity i.e., even or odd and y i and y j have the same parity. There are possible pairs of parities each x i, y i could have namely even, even, even, odd, odd, even, and odd, odd, so by the Pigeonhole Principle two x i, y i pairs must have the same parities: there must exist x i, y i and x j, y j such that x i and x j have the same parity, and y i and y j have the same parity. Then the midpoint of the line between these two points has integer coordinates. 6.2: 2: There are 5 possible remainders when an integer is divided by 5: 0,, 2, 3, and. Thus there are possible pairs of remainders a mod 5, b mod 5. We therefore require 26 ordered pairs: by the Pigeonhole Principle, some two pairs a, b and a 2, b 2 will have the same pair of remainders, so a mod 5 a 2 mod 5 and b mod 5 b 2 mod : 2: There are 5 choices for each selection, so there are 5 5 ways. 6.5: : Each day there are 6 choices, so there are 6 7 ways. 6.5: 8: This is the number of 2-combinations of 2 things with repetition. There are ways because repetition is allowed.

2 6.5: 2: This is the number of 20-combinations of 5 things with repetition, which is : : This is the number of 7-combinations of things with repetition. Why? The number of 7-combinations of things can be thought of as the number of ways to put 7 indistinguishable balls into bins. The number of balls in the bins correspond to the values of the x i and add up to 7. There are ways. 20! 6.5: 8: There are such strings. Why? There are 20! different ways to permute 2!!3!!2!3!2!3! the digits, but this overcounts: there are 2! ways to reorder the 0 s to the same effect,! ways to reorder the s without changing the word, etc. Dividing by the number of times we overcount gives the above expression. 6.5: 20: We can write this as x + x 2 + x 3 + x, where x is defined as x + x 2 + x 3. Then we re looking for the number of -combinations of things with repetition, which is : 22: This is the number of 2-combinations of 6 things, : 2: There are 5!!2!3!!5! ways. This is a similar answer as that to problem 8, despite looking like a different sort of problem. You can see the correspondence as follows: imagine we just put the 5 objects in a row and beneath each object we write the index of the box we re going to put it in. Then really this problem is asking for the number of ways to make a string of 5 digits out of one, 2 2 s, 3 3 s, s, and 5 5 s, just as in problem 8 but with different numbers. 6.5: 26: The positive integers less than are those consisting of 6 digits from 0-9 in a row with leading 0 s allowed. Call the 6 digits x, x 2, x 3, x, x 5, and x 6. Then we need x + x 2 + x 3 + x + x 5 + x 6 3 and some x i 9. There are 6 ways to pick which x i 9. The remaining 5 x i s must then sum to 3 9 ; this is the number of -combinations of 5 things, which is +5 8, giving a total of 6 8 ways. But wait, didn t we just overcount? If more than one x i is simultaneously equal to 9, then this argument will count that number more than once. Fortunately that can t happen because the sum of the digits is only 3 < : 3: In this magic word there are 5 A s, 2 B s, C, D, and 2 R s, so the number of permutations of its letters is! 5!2!!!2!. 6.5: 38: In this problem we assume the journals are distinguishable, or else the problem is easy the answers are both. a There are 0! 0!0!0!0! such ways. b Our answer to a now overcounts by a factor of! the number of reorderings of the 2

3 boxes so there are! 0! 0!0!0!0! ways. 6.5: 0: This is the same as the number of strings of length 6 consisting of x s, 3 y s, 5 z s, and w s think of an x as meaning take one step in the positive x direction, and similarly for y, z, and w. There are then 6!!3!5!! ways. 6.5: 2: Let s assume that each player gets the same number of cards, 3 as happens in 52! bridge. There are then ways. Why? There are 52! ways to permute the cards; 3!3!3!3! imagine the first 3 going to player, the next 3 to player 2, etc. But this overcounts by the number of ways to permute the cards player gets 3! and that player 2 gets 3! and that player 3 gets 3! and that player gets 3!. Dividing by this overcounting factor gives the above answer. 6.5: : a This is the number of 2-combinations of things, which is b Let s place the books one-by-one, inserting them between previously placed books or at the start/end of a shelf. When we place the first book, there are only places to put it. When we place the second book, there are now 5 spots: it could go on one of the empty shelves, or it could go on the same shelf as the previous book, either to its right or to its left. Similarly there are 6 places to put the third book, and so on. Thus there are ! ways, the product of the first 2 3! integers starting with. What does this have to do with r-combinations with repetition? Suppose we take a permutation of the integers -2 together with 3 bars, such as 7,3,9,6,2, 2,3 5,,8. We can think of this as specifying an ordering of books on shelves, where the bars separate the shelves, in order. How many such permutations are there? If the bars were distinguishable there would be 2 + 3! 5!, but this overcounts by a factor of 3! due to reordering the bars. Thus there are 5! ways. This is the same as the proof of 3! the formula for r-combinations with repetition, except that our non-bar elements are distinguishable instead of indistinguishable. 6.5: 6: As indicated in the hint, think of the books we choose as bars and the books we don t choose as stars. Then we we re counting all the permutations of these 5 bars and 7 stars such that there is at least one star between every pair of bars. Imagine the 5 bars are placed first:. Now place one star between every consecutive pair of bars:. There are now 3 stars left that we have to distribute amongst 6 spaces; this is the number of 3-combinations of 6 things with repetition, which is : 50: This is S5, 3 + S5, 2 + S5, remember that Sn, k the Stirling number of the second kind is the number of ways to put n distinguishable things into k indistinguishable boxes such that each box has at least one thing in it. 3

4 Let s count this by brute force. There are 5! 5 ways to put the objects in 2!!2!2! boxes such that the numbers of items in boxes are, 2, and 2, in some order. Why? If 5! the boxes were distinguishable there would be ways, but this overcounts by a factor!2!2! of 2! due to the size-2 boxes being indistinguishable. Using the same logic, we get this table: Number of elements per box,2,2,,3 2,3, 5 Count 5! 5! 5! 5! 2!!2!2! 2!!!3! 2!3!!! Numerical answer Thus there are ways. Also, from the first two entries we can see that S5, , from the next two entries that S5, , and from the last entry that S5,. 6.5: 65: Imagine writing out x + y + z 0 as x + y + zx + y + zx + y + z...x + y + z. When we expand this out every term in the expansion comes from picking either an x or a y or a z in each term. To get a coefficient of x 3 y 2 z 5, we must pick x from 3 of the terms, y from 2 of the terms, and z from 5 of the terms. The number of ways to do this same 0! as the number of strings of length 0 with 3 x s, 2 y s, and 5 z s is. This is why this 3!2!5! expression is called a multinomial coefficient, in the same sense that n k n! is called k!n k! a binomial coefficient. 6.5: 66: Every term is of the form x a y b z c where a + b + c 00 and a, b, c 0. Similarly, every such choice of a, b, and c constitutes a term in this expansion. The number of such choices is the number of 00-combinations of 3 things, which is : 6: There are 3 hearts and an additional 3 aces remember not to overcount the ace of hearts, so there are 6 cards that are either an ace or a heart. Thus the probability of drawing one of these cards is : 8: There are 5 hands containing this card out of all card hands, so the probability is !!7!. 52! 52 5! 7! 5 7.: 2: There are ways to pick which ace it contains, and there are 8 cards remaining all the non-aces from which to pick more cards, so the probability is : 6: In each of the suits there are 3 5 5! 5! such hands. Thus the probability is 3 7.: 8: Suppose first that aces are considered to be the same as. In each of the suits there are 9 such hands, the runs of 5 cards starting between and ending at 5 and 9 and ending at 3K. There are similarly 9 straight flushes per suit if you consider aces to be high, this time starting between 2 and 0. Therefore the probability of getting a

5 straight flush is 9. If aces are allowed to be used as either a low or high card depending 52 5 on the rest of the hand there are 0 straight flushes per suit, so a probability of : 22: There are 00 such integers, 33 of which are divisible by 3 the integers of the form 3i as i ranges from to 33, so the probability is : 2d: There are 8 6 possible combinations and only is a winner, so the probability is, or approximately in 2 million : 26b: Out of the 8 integers 2 are incorrect, so there are 2 6 ways to pick entirely the wrong numbers. Thus the probability of doing so is So 3% of the time 8 6 none of your numbers will be right. 7.: 30: Regardless of what the correct numbers are, there are 6 ways to pick which number we don t match and ways to pick which incorrect number we choose instead. The probability of this happening is then 6 3, or about in 9, : 3: a The probability of Bo winning the first prize is. The probability of Colleen winning 50 the next prize is then since Bo cannot win it, meaning there are 9 potential 9 winners, and so on. The probability is then b The probability of Bo winning the first prize is, the probability of Colleen winning 50 the next prize is, and so on. Thus the probability is : 36: When rolling 2 dice, there are possible rolls and 5 ways to get a sum of 8 let s think of the dice as being distinguishable, say one is blue and the other is red: 2+6, 3+5, +, 5+3, 6+2. So the probability is When rolling 3 dice, there are possible rolls. How many ways to get a sum of 8? Let s count them out to speed up the process we won t write out all the possible ways of permuting the summands e.g. 6++ versus +6+ versus ++6: Sums to 8 permutations Number of equivalent permutations Thus there are ways there are other ways to count this, such as putting 8 indistinguishable balls into 3 bins such that each bin gets at least, and then subtracting off all cases where a bin gets more than 6, so the probability of getting a sum of 8 is So it s more likely we ll get a sum of 8 when rolling 2 dice. This 26 shouldn t be very surprising: when rolling 2 dice we d expect to get 7 on average, which 5

6 is pretty close to 8, while rolling 3 dice we d expect to get 0.5 on average, which is much farther away. 7.: 38: a. We have pe 2, pe 2 2, and pe E ; since pe pe 2 pe E 2, the events are independent. b. We have pe 2. There are 2 ways for E 2 to happen HHT and THH so pe For both E and E 2 to happen we must get THH, so pe E 2 8. Since pe pe 2 pe E 2, the events are independent. c. We have pe and pe as before. Now pe 8 E 2 0, since if the second coin comes up tails it s impossible to have 2 heads in a row. Since pe pe 2 pe E 2, these events are not independent. 7.: 0: There is a in chance that our initial guess was correct, so by not changing doors our probability of winning is. Suppose we instead decide to change doors. After the host opens a losing door, 3 doors remain, one of which is correct. In the event that our initial guess was correct probability, changing doors gives us a probability 0 of winning. But if our initial guess was wrong probability 3, out of the 2 remaining doors excluding the door the host opened and excluding our initial guess one is correct, so we have a 2 chance of winning. Thus our chance of winning is Since 3 >, changing doors is the better strategy. 8 7.: : a Let s find the probability of *not* rolling at least one 6. The probability of not getting a 6 on our first roll is 5, of not getting a 6 on our second roll is 5, etc. So the probability 6 6 of *not* getting at least one 6 is 5, 6 and so the probability of getting at least one 6 is b The probability of *not* getting a double 6 on our first roll is, of not getting 36 a double 6 on our second roll is, etc. So the probability of not getting any 36 double 6s is 36 2, and so the probability of getting at least one double 6 is So the probability of getting a double 6 is slightly under half that is, it s slightly more likely we will not get a double 6 than that we will. c It s more likely that a 6 will come up at least once when rolling a die times happens 5.8% of the time than that we ll get at least one double 6 out of 2 rolls of two dice happens 9.% of the time. 6