# 94 Counting Solutions for Chapter 3. Section 3.2

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5 98 Counting Solution: By the subtraction principle combined with the answer to the first part of this problem, the answer is 2 10 ( 10 ( A 5-card poker hand is called a flush if all cards are the same suit. How many different flushes are there? Solution: There are ( 13 5 = card hands that are all hearts. Similarly, there are ( 13 5 = card hands that are all diamonds, or all clubs, or all spades. By the addition principle, there are then = 5148 flushes. Section Write out Row 11 of Pascal s triangle. Answer: Use the binomial theorem to find the coefficient of x 8 in (x Answer: According to the binomial theorem, the coefficient of x 8 y 5 in (x y 13 is ( 13 5 x 8 y 5 = 1287x 8 y 5. Now plug in y = 2 to get the final answer of 41184x Use the binomial theorem to show ( n nk k=0 = 2 n. Hint: Observe that 2 n = (11 n. Now use the binomial theorem to work out (x y n and plug in x = 1 and y = Use the binomial theorem to show n k=0 3k ( n k = 4 n. Hint: Observe that 4 n = (1 3 n. Now look at the hint for the previous problem. 9. Use the binomial theorem to show ( n 0 ( n1 ( n2 ( n3 ( n4 ( n5... ± ( nn = 0. Hint: Observe that 0 = 0 n = (1 (1 n. Now use the binomial theorem. 11. Use the binomial theorem to show 9 n = n ( k=0 (1k n k 10 nk. Hint: Observe that 9 n = (10 (1 n. Now use the binomial theorem. 13. Assume n 3. Then ( n ( 3 = n1 ( 3 n1 ( 2 = n2 ( 3 n2 ( 2 n1 ( 2 = = 2 ( 2 3 ( 2 n1 2. Section At a certain university 523 of the seniors are history majors or math majors (or both. There are 100 senior math majors, and 33 seniors are majoring in both history and math. How many seniors are majoring in history? Solution: Let A be the set of senior math majors and B be the set of senior history majors. From A B = A B A B we get 523 = 100 B 33, so B = = 456. There are 456 history majors. 3. How many 4-digit positive integers are there that are even or contain no 0 s? Solution: Let A be the set of 4-digit even positive integers, and let B be the set of 4-digit positive integers that contain no 0 s. We seek A B. By the multiplication principle A = = (Note the first digit cannot be 0 and the last digit must be even. Also B = = Further, A B consists of all even 4-digit integers that have no 0 s. It follows that A B = = Then the answer to our question is A B = A B A B = = 8145.

6 Solutions for Chapter How many 7-digit binary strings begin in 1 or end in 1 or have exactly four 1 s? Solution: Let A be the set of such strings that begin in 1. Let B be the set of such strings that end in 1. Let C be the set of such strings that have exactly four 1 s. Then the answer to our question is A B C. Using Equation (3.5 to compute this number, we have A B C = A B C A B A C B C A B C = ( ( 6 3 ( 63 ( 52 = = This problem concerns 4-card hands dealt off of a standard 52-card deck. How many 4-card hands are there for which all four cards are of the same suit or all four cards are red? Solution: Let A be the set of 4-card hands for which all four cards are of the same suit. Let B be the set of 4-card hands for which all four cards are red. Then A B is the set of 4-card hands for which the four cards are either all hearts or all diamonds. The answer to our question is A B = A B A B = 4 ( 13 4 ( ( 134 = 2 ( 13 4 ( 264 = = A 4-letter list is made from the letters L,I,S,T,E,D according to the following rule: Repetition is allowed, and the first two letters on the list are vowels or the list ends in D. How many such lists are possible? Solution: Let A be the set of such lists for which the first two letters are vowels, so A = = 144. Let B be the set of such lists that end in D, so B = = 216. Then A B is the set of such lists for which the first two entries are vowels and the list ends in D. Thus A B = = 24. The answer to our question is A B = A B A B = = How many 7-digit numbers are even or have exactly three digits equal to 0? Solution: Let A be the set of 7-digit numbers that are even. By the multiplication principle, A = = 4,500,000. Let B be the set of 7-digit numbers that have exactly three digits equal to 0. Then B = 9 ( (First digit is anything but 0. Then choose 3 of 6 of the remaining places in the number for the 0 s. Finally the remaining 3 places can be anything but 0. Note A B is the set of 7-digit numbers that are even and contain exactly three 0 s. We can compute A B with the addition principle, by dividing A B into two parts: the even 7-digit numbers with three digits 0 and the last digit is not 0, and the even 7-digit numbers with three digits 0 and the last digit is 0. The first part has 9 (5 ( elements. The second part has elements. Thus A B = 9 (5 ( By the inclusion-exclusion formula, the answer to our question is A B = A B A B = ( ( ( 5 2 = 4,405, How many 8-digit binary strings end in 1 or have exactly four 1 s? Solution: Let A be the set of strings that end in 1. By the multiplication principle A = 2 7. Let B be the number of strings with exactly four 1 s. Then B = ( 8 4 because we can make such a string by choosing 4 of 8 spots for the 1 s and filling the remaining spots with 0 s. Then A B is the set of strings that end with 1 and have exactly four 1 s. Note that A B = ( 7 4 (make the last entry a 1 and choose 3 of the remaining 7 spots for 1 s. By the inclusion-exclusion

7 100 Counting formula, the number 8-digit binary strings that end in 1 or have exactly four 1 s is A B = A B A B = 2 7 ( 8 4 ( 7 3 = How many 10-digit binary strings begin in 1 or end in 1? Solution: Let A be the set of strings that begin with 1. By the multiplication principle A = 2 9. Let B be the number of strings that end with 1. By the multiplication principle B = 2 9. Then A B is the set of strings that begin and end with 1. By the multiplication principle A B = 2 8. By the inclusionexclusion formula, the number 10-digit binary strings begin in 1 or end in 1 is A B = A B A B = = 768. Section A bag contains 20 identical red balls, 20 identical blue balls, 20 identical green balls, and one white ball. You reach in and grab 15 balls. How many different outcomes are possible? Solution: First we count the number of outcomes that don t have a white marble. Modeling this with stars and bars, we are looking at length-17 lists of the form red blue green, where there are 15 stars and two bars. Therefore there are ( outcomes without the white ball. Next we count the outcomes that do have the white ball. Then there are 14 remaining balls in the grab. In counting the ways that they can be selected we can use the same stars-and-bars model above, but this time the list is of length 16 and has 14 stars. There are ( outcomes. Finally, by the addition principle, the answer to the question is ( 17 ( = In how many ways can you place 20 identical balls into five different boxes? Solution: Let s model this with stars and bars. Doing this we get a list of length 24 with 20 stars and 4 bars, where the first grouping of stars has as many stars as balls in Box 1, the second grouping has as many stars as balls in Box 2, and so on. Box 1 Box 2 Box 3 Box 4 Box 5, The number of ways to place 20 balls in the five boxes equals the number of such lists, which is ( = 10, A bag contains 50 pennies, 50 nickels, 50 dimes and 50 quarters. You reach in and grab 30 coins. How many different outcomes are possible? Solution: The stars-and-bars model is pennies so there are ( = 5456 outcomes. nickels dimes quarters,

8 Solutions for Chapter How many integer solutions does the equation w x y z = 100 have if w 4, x 2, y 0 and z 0? Solution: Imagine a bag containing red, blue, green and white balls. Each solution of the equation corresponds to an outcome in grabbing 100 balls from the bag, where there are w 4 red balls, x 2 blue balls, y 0 green balls and z 0 white balls. Pre-select 4 red balls and 2 blue balls. Now the remaining 92 balls are selected. We can calculate the number of ways that this selection can be made with stars and bars, where there are 92 stars and 3 bars. red blue green white, The number of outcomes is thus ( = 138, How many permutations are there of the letters in the word "TENNESSEE"? Solution: By Fact 3.8, the answer is 9! 4!2!2! = 3, You roll a dice six times in a row. How many possible outcomes are there that have two 1 s three 5 s and one 6? Solution: This is the number of permutations of the word. By 6! Fact 3.8, the answer is 2!3!1! = In how many ways can you place 15 identical balls into 20 different boxes if each box can hold at most one ball? Solution: Regard each such distribution as a binary string of length 20, where there is a 1 in the ith position precisely if the ith box contains a ball (and zeros elsewhere. The answer is the number of permutations of such a string, which by Fact 3.8 is 20! 15!5! = 15,504. Alternatively, the answer is the number of ways to choose 15 positions out of 20, which is ( = 15, How many numbers between 10,000 and 99,999 contain one or more of the digits 3, 4 and 8, but no others? Solution: First count the numbers that have three 3 s, one 4, and one 8, like 33,348. By Fact 3.8, the number of permutations of this is 5! 3!1!1! = 20. By the same reasoning there are 20 numbers that contain three 4 s, one 3, and one 8, and 20 numbers that contain three 8 s, one 3, and one 4. Next, consider the numbers that have two 3 s, two 4 s and one 8, like 33,448. By Fact 3.8, the number of permutations of this is 5! 2!2!1! = 30. By the same reasoning there are 30 numbers that contain two 3 s, two 8 s and one 4, and 30 numbers that contain two 4 s, two 8 s and one 3. This exhausts all possibilities. By the addition principle the answer is = 150. Section 3.10 Exercises 1. Show that if 6 integers are chosen at random, at least two will have the same remainder when divided by 5.

9 102 Counting Solution: Pick six integers n 1, n 2, n 3, n 4, n 5 and n 6 at random. Imagine five boxes, labeled Box 0, Box 1, Box 2, Box 3, Box 4. Each of the picked integers has a remainder when divided by 5, and that remainder is 0, 1, 2, 3 or 4. For each n i, let r i be its remainder when divided by 5. Put n i in Box r i. We have now put six numbers in five boxes, so by the pigeonhole principle one of the boxes has two or more of the picked numbers in it. Those two numbers have the same remainder when divided by What is the fewest number of times you must roll a six-sided dice before you can be assured that 10 or more of the rolls resulted in the same number? Solution: Imagine six boxes, labeled 1 through 6. Every time you roll a, put an object in Box 1. Every time you roll a, put an object in Box 2, etc. After n rolls, the division principle says that one box contains n 6 objects, and this means you rolled the same number n 6 times. We seek the smallest n for which n This is the smallest n for which n 6 > 9, that is n > 9 6 = 36. Thus the answer is n = 37. You need to roll the dice 37 times. 5. Prove that any set of 7 distinct natural numbers contains a pair of numbers whose sum or difference is divisible by 10. Solution: Let S = {a 1, a 2, a 3, a 4, a 5, a 6, a 7 } be any set of 7 natural numbers. Let s say that a 1 < a 2 < a 3 < < a 7. Consider the set A = {a 1 a 2, a 1 a 3, a 1 a 4, a 1 a 5, a 1 a 6, a 1 a 7, a 1 a 2, a 1 a 3, a 1 a 4, a 1 a 5, a 1 a 6, a 1 a 7 } Thus A = 12. Now imagine 10 boxes numbered 0,1,2,...,9. For each number a 1 ± a i A, put it in the box whose number is the one s digit of a 1 ± a i. (For example, if a 1 ± a i = 4, put it in Box 4. If a 1 ± a i = 8, put it in Box 8, etc. Now we have placed the 12 numbers in A into 10 boxes, so the pigeonhole principle says at least one box contains two elements a 1 ± a i and a 1 ± a j from A. This means the last digit of a 1 ± a i is the same as the last digit of a 1 ± a j. Thus the last digit of the difference (a 1 ± a i (a 1 ± a j = ±a i ± a j is 0. Hence ±a i ± a j is a sum or difference of elements of S that is divisible by 10.

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