Some Review Problems for Exam 3: Solutions

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1 Math 3355 Fall 018 Some Review Problems for Exam 3: Solutions I thought I d start by reviewing some counting formulas. Counting the Complement: Given a set U (the universe for the problem), if you want to now how many things in U have some property, an indirect approach is to count how many things don t have that property instead. The formula: The number of things that do = U the number that don t. Said differently, if A is a subset of U, and Ā is the complement of A in U, then A = U Ā. Inclusion-Exclusion: A B = A + B A B. The binomial coefficient ( n ) counts: -element subsets of a set of size n, bit strings of length n having exactly ones, committees of size from a group of n people. Permutations: P (n, ) = n(n 1) (n + 1) = n! (n )! counts the number of ways to arrange things from n types of things, if repetition is not allowed. If repetition IS allowed, the formula is easier, in this case, there are n possible arrangements. The other important idea is a combinatorial proof of an identity. We establish the identity by counting something in two different ways. Typically, one side of the identity is simpler than the other. The simpler side usually has a direct count. The tricier side often involves breaing up the count into cases. Here are the review question solutions: 1. The English alphabet contains 1 consonants and five vowels. How many strings of four lowercase letters of the English alphabet contain: (a) A vowel in position? Solution: This is a product rule: = = On an exam, or better, are fine. In fact, I would say that is a better answer than because it is easier to grade and shows some of the ideas going into getting the count. (b) Vowels in positions and 3? Solution: Same idea as in (a): = 5 6, or

2 (c) A vowel in position or in position 3? Solution: This is harder. One way is to count the complement: Subtract off the number of strings that don t have a vowel in either position from the total: U Ā = = An alternative is to use inclusion-exclusion: A B = A + B A B. In (a) we said strings have a vowel in position, and the same number have a vowel in position 3. In (b), we said 5 6 have vowels in both positions. The total is = , as before. (d) At least one consonant? Solution: This is a classic complement problem: U Ā = = (e) Exactly one vowel? Solution: I would count this by the following multistep approach: Figure out where you want the vowel to be, decide which vowel to put there, and then fill the remaining 3 positions with consonants. This leads to the answer = = (f) No vowels in consecutive positions? ( have is OK, beat is not) Solution: This is hard. One approach is to use cases: How many vowels are there in the string? The possibilities are no vowels, one, or two vowels. No vowels gives 1 4, one vowel gives (the answer to part (e).) With two vowels, we have to figure out how many ways to place them. The possibilities are vcvc, vccv, cvcv, for a total of three ways. Next, we pic two vowels and two consonants. The count for this case is All totaled, we have = Suppose Minnesota Pic 3 became Minnesota Pic 4 instead (you select four numbers, with order counting, each number between 0 and 9, inclusive). (a) How many combinations are possible? Solution: 10 4 = 10, 000 Page

3 (b) How many combinations don t contain a 5? Solution: 9 4 = 6561 (c) How many combinations contain a 5? Solution: Complement problem: (answer to (a) - answer to (b)) = 10, = (d) How many combinations contain exactly two 5 s? Solution: There are ( 4 ) = 6 ways to place the fives, and 9 9 ways to fill the other position, so the total is 6 81 = 486. (e) How many combinations contain at least two 5 s? Solution: Another complement problem, but care must be used. The complement has two cases: no fives or one five. If no fives, we get 6561, as in (b). If one five, we have = 916 ways. Our answer is 10, = 53 combinations. An alternative is to add up cases with two, three, or four fives. There are 486 ways with two fives, ( 4 3) 9 ways with three fives, and one way with four fives, for a total of = How many bit strings of length 6 are there containing: (a) Exactly two 1 s? Solution: ( ) 6 = 6 5 = 15. (b) At least two 1 s? Solution: Count the complement: 6 ( 6 1) ( 6 0) = = 57. Page 3

4 (c) Exactly two 1 s, with the 1 s in nonconsecutive positions? Solution: Fastest might be to just list all the possibilities: for a total of , , , , , , , , , An alternative is to count the complement: Of the 15 strings with two ones, , , , , have the ones in consecutive positions, giving a total of 15-5 = 10. (d) At least two 1 s with no 1 s in consecutive positions? Solution: Again, it might be fastest to just count the possibilities. We have done the case with exactly two ones. If we have three ones, there are four more cases: , , , , and you can t do it with more than three ones, so we have a total of = 14 cases. 4. (a) Find the coefficient of x 17 in (x 3) ( ) 40 Solution: By the Binomial Theorem, (x 3) 40 = (x) 40 ( 3). We want the stuff that multiplies x 17. This means we want 40 = 17, or = 3. The answer we see is ( ) ( 3) 3, or ( 40 3) ( ) (b) Evaluate 5. =0 Solution: By a direct application of the Binomial Theorem, this will be ( + 5) = 7. ( ) (c) Evaluate. =0 Solution: This is also a Binomial Theorem question, with x = 1, y = 1, giving (1 + 1) =. =0 Page 4

5 ( ) (d) Evaluate =0 ( ( ) ( ) ( ) ( )) or Solution: One way, as mentioned in class, is that n(1 + x) n 1 = d dx (1 + x)n = d dx =0 n =0 ( ) n x = n =0 ( ) n x 1. ( ) Setting x = 1 and n = gives = (1 + 1) 49 = 49. This ( ) could( also be ) done without calculus. A binomial ( ) coefficient ( identity ) is n n 1 49 = n. In the given problem, this says =. If we 1 1 do things without summation notation (so the pattern might be more obvious) this goes as follows: ( ) ( ) ( ) ( ) ( ) ( ) ( ) = ( ) ( ) ( ) = (( ) ( ) ( )) = = Consider the word SIZIGIES. (a) How many arrangements are there of the word? Solution: There are 8 letters: an E, a G, three I s, two S s and a Z. The number 8! of arrangements is 1!1!3!!1! = 8!, or !! (b) How many arrangements have the three I s together? Solution: There are 6 ways to place the I s (the first I can be in any of positions 1 through 6). This leaves 5 additional positions. We still have to pay attention 5! to identical letters (the two S s) so the total is 6 1!1!!1! = 6! = 360. Page 5

6 (c) How many don t have the three I s together? Solution: We count the complement, the answer being the difference of the previous two answers: = (d) How many don t have any I s together? Solution: Another hard problem. One could enumerate the possible positions of the I s: In positions 1, 3, 5 or 1, 3, 6 or.... I get 0 possible ways to place the I s. One way to be systematic: First place an I in position 1. If the second I is in position 3, there are 4 places for the next I. If the second I is in position 4, there are 3 ways and so on, leading to = 10 ways to have an I in position 1. You might now be able to guess there are = 6 ways for the first I to be in position, + 1 = 3 ways for the first I to be in position 3, and 1 way if it is in position 4 for a total of = 0 ways. We still 5! have to arrange the other letters. As in part (b), this happens in 1!1!!1! = 60 ways. The total is 0 60 = This problem deals with solutions to x + y + z = in integers. (a) How many solutions are there with x, y, z nonegative? Solution: ( The ) simplest ( ) version of the problem. We have n =, = 3 so we get = = 136 ways. 3 1 (b) How many solutions have x 5, y 10, z 15? Solution: If you thin of balls in boxes, then we need at least 5 balls in the first box, 10 in the second and 15 in the third. First we put 30 of the balls in the boxes so as to get the required minimums. Then we( put the remaining ) ( 0) palls in boxes with no additional restrictions. The total is = = ways. (c) How many solutions have x 0, y 0, 0 z 30? Solution: We count the complement: ( If the ) third( condition ) were false, then z 31 which can happen in = = 10 ways. We get ( ) ( ) = = Page 6

7 (d) How many solutions have 0 x, y, z 30? (All variables are between 0 and 30.) Solution: This is similar to c in that we count the complement. However, the complement to all three variables being small is that at least one of them be too big. Fortunately, we can t have more than one large variable (if x > 30 and y > 30 then we would have more than for the total). The count goes lie this: ( Pic) the variable to be too big: 3 ways. Then do the count as in c. We 1 have = 10 ways a variable could be too big times 3 variables for 630 bad cases. The total is = The US Senate has 100 members, two from each of the states. (a) How many committees can be formed from the 100 senators? Solution: This question ass how many subsets a 100-element set can have. The answer: 100. (b) How many committees are there with exactly one senator from each state? Solution: For each state, pic a senator. The answer:. (c) How many committees of size 5 are there? ( ) 100 Solution: We need five people from the 100 senators, so. 5 (d) How many committees of size 5 are there if you are not allowed two senators from the same state? Solution: Select ( ) which of the five states to use, and then pic a senator from each state: 5. 5 (e) How many committees of size 5 are there if you are not allowed two senators from the same state, but one of the senators must be from Minnesota? Solution: Pic a Minnesota senator, then pic four of the ( remaining ) ( states, ) and last, a senator from each of those states. The count is 4 = Page 7

8 8. Give a combinatorial proof for each of the following. (a) n 3 = (n 1) 3 +3(n 1) +3(n 1)+1 Hint: Count strings xyz where each of x, y, z is an integer between 1 and n. For the right hand side, brea into cases based on how many 1 s there are in the string. Solution: We count in two ways, as suggested by the hint. The direct approach is that each of x, y, z has n choices for a total of n 3 possibilities. For the more complicated count, we consider how many 1 s there are in the string, with four cases: no ones, one 1, two 1 s, three 1 s or four 1 s. With no ones, there are n 1 choices for each of x, y, x leading to a count of (n 1) 3. If there is one 1, pic one of x, y, z to be 1, and the others have n 1 choices for a total of 3(n 1). If we have two 1 s, select which two (three ways), and the third variable has n 1 choices, giving 3(n 1), and there is only one string, 1 1 1, with three 1 s. The total from this method is (n 1) 3 +3(n 1) +3(n 1)+1. (b) Since both methods are legitimate approaches, they must give the same answer, so n 3 = (n 1) 3 + 3(n 1) + 3(n 1) + 1. ( ) ( ) 3n n = 3 + 3n Hint: Form committees of size from a group of n administrators, n teachers and n students. Solution: If we ignore class (administrator vs teacher vs student), then we have a total of 3n( people ) and we want a committee of size. Such committees can 3n be piced in ways, giving the Left Hand Side of the expression. For the Right Hand Side, consider cases based on the maeup of the committee. There are a total of 6 cases: If both( committee ) members are administrator, then with n n to choose from we have all Seniors committees. Similarly, we have ( ) ( ) n n all teacher or all student committees, leading to 3 committees when both people are of the type. Three other cases involve mixed type committees. We could have administrator-teacher, administrator-student, or teacher-student selections. In each of these three cases, there are n ways to pic one person and n ways to pic the other since in each case there are n to choose from. The total from( these ) cases is 3 n n = 3n. All totaled, counting by committee mae n up gives 3 + 3n committees, which must give the same answer as the first ( ) ( ) 3n n count so = 3 + 3n. Page 8

9 9. Which nonnegative integers can be written as sums of 5 s and 6 s? Prove your answer is correct, both with regular induction and by strong induction. Solution: We start with a list of possible numbers: 5, 6, 10, 11, 1, 15, 16, 17, 18, 0, 1,, 3, 4, 5, 6,.... The first 9 numbers on the list are incidental or accidents of mathematics. What is to be proved is that all integers from 0 on can be written as sums of 5 s and 6 s. First, the strong induction proof. To be careful, we again verify the base cases. In this problem, there are five of them (getting 5 in a row). They are: 0 = , 1 = , = 5 + 6, 3 = , and 4 = Let P (n) be the statement that n is the sum of 5 s and 6 s. Having verified the base case, the inductive hypothesis is that P () is true for all with 0 n, where n is some integer with n 4. The goal is to prove that P (n + 1) is also true. That is, we have to show that n + 1 is the sum of 5 s and 6 s. Intuitively, the reason things wor is that once we have five things in a row, we should be able to just at 5 s to previous things. Since this is a class related to proving things, we should be more rigorous. What should we add 5 to in order to get n + 1? The answer is n 4. So here is the proof that n + 1 is the sum of 5 s and 6 s: Since n 4, n 4 0. By our inductive hypothesis, anything between 0 and n can be written as the sum of 5 s and 6 s, so there are integers j and for which n 4 = 5j + 6. Adding 5 to this, n + 1 = 5(j + 1) + 6, so n + 1 is also the sum of 5 s and 6 s. I thin the standard induction argument is tricier. We need to be able to get from n to n + 1 rather than from n 4 to n + 1. To increase by 1, we loo for ways to convert 5 s to 6 s so as to increase the total by 1, and ways to convert 6 s to 5 s to do the same. It is easy to convert 5 s to 6 s, of course: If we have even a single 5, replacing it by a 6 increases the sum by 1. The other direction is slightly harder: we can trade four 6 s for five 5 s to increase the sum by 1. Said differently, suppose that n = 5m + 6 for some integers m and. Then n + 1 = 5(m 1) + 6( + 1), and n + 1 = 5(m + 5) + 6( 4). This means we can go from n to n + 1 if we have at least one 5 or four 6 s. The cases where we can t get from n to n + 1 must be when we have fewer than one 5 and fewer than four 6 s. But = 18, so if n > 18 and n is the sum of 5 s and 6 s, then we must have at least one 5 or at least four 6 s. What this means is that as soon as we find an n > 18 which is a sum of 5 s and 6 s, all larger numbers should be sums of 5 s and 6 s as well. The smallest number n > 18 which is the sum of 5 s and 6 s is n = 0. (If 19 were the sum of 5 s and 6 s we could have started there, but it isn t.) With all this preparation, here is the actual proof by regular induction. We prove that each n 0 is the sum of 5 s and 6 s. The base case is n = 0, and since Page 9

10 0 = , the base case is true. So let n 0, and suppose that n is the sum of 5 s and 6 s, say n = 5m + 6. Since n > 0, either m 1 or 4. If m 1, then n + 1 = 5(m 1) + 6( + 1). If 4, then n + 1 = 5(m + 5) + 6( 4). Either way, n + 1 is the sum of 5 s and 6 s. By the principle of (regular) induction, all numbers n 0 are sums of 5 s and 6 s. 10. Here are some more strong induction problems. (a) Show that for the Fibonacci numbers, the pattern is EVEN, ODD, ODD. That is the sequence is 0, 1, 1,, 3, 5, 8, 13, 1, 34, etc. In terms of formulas, F 3n should always be even, F 3n+1 and F 3n+ should be odd. Solution: With 0, 1 1, we see the pattern starts EVEN, ODD, ODD. By way of induction, suppose this pattern persists through the first 3n Fibonacci numbers. That is, we may assume F 3n is even, F 3n+1 and F 3n+ are odd. We show that the next three Fibonacci numbers also follow the pattern. Let F 3n = j, F 3n+1 = + 1 and F 3n+ = m + 1. Then F 3n+3 = F 3n+ + F 3n+1 = m = (m + + 1). Thus, F 3n+3 is even. Also, F 3(n+1)+1 = F 3n+3 + F 3n+ = m + n + + m + 1 = (m + n + 1) + 1, F 3(n+1)+ = F 3n+4 + F 3n+3 = (m + n + 1) (m + n + 1) = (3m + n + ) + 1, so F 3(n+1)+1 and F 3(n+1)+ are both odd. Thus, the pattern of EVEN, ODD, ODD continues to these three Fibonacci numbers as well. By the Principle of Mathematical Induction, this pattern continues indefinitely. (b) In class, I showed that a 1 n candy bar taes n 1 breas at dividers to get n 1 1 pieces, no matter how you divide the bar. What if we have an m n bar, now how many breas are needed? Solution: If we had a 3 bar, we could brea it into two 1 3 bars, and each bar would need two additional breads for a total of = 5 breas. This suggests mn 1 breas are needed. We prove this by induction, the result being true for small m and n. Given an m n bar, the first brea must be either vertical or horizontal. If vertical at position, we have two bars of size m and m (n ). By inductive hypothesis, these tae m 1 and m(n ) 1 additional breas to get to 1 1 bars, for a total of 1 + m 1 + m(n ) 1 = m + mn m 1 = mn 1 breas. Using a horizontal brea at position is similar. By strong induction, any m n bar requires mn 1 breas. Page 10

11 (c) Prove that n 3 n is always divisible by 6. One way: Do six cases in a row. Now assuming (n 5) 3 (n 5) is divisible by 6, show (n + 1) 3 (n + 1) also is. To help, write m = n 5. You now m 3 m is divisible by 6, and n + 1 = m + 6 so you need to show (m + 6) 3 (m + 6) is divisible by 6. Solution: I thin this is ind of cute. I did not say where n should start, let s start at n = 0. For 0 n 5 we get values of 0, 0, 6, 4, 60 and 10 for n 3 n, and we see that all of these are divisible by 6. By way of induction, suppose that 3 is divisible by 6 for all with 0 n, for some n 5. We want to show that (n + 1) 3 (n + 1) is divisible by 6. We do so by letting m = n 5 so n + 1 = m + 6. We have (n + 1) 3 (n + 1) = (m + 6) 3 (m + 6) = m m + 108m + 16 m 6 = (m 3 m) + 6(3m + 18m + 35). Since n 5, m = n 5 0. More to the point, 0 m n so by inductive hypothesis, m 3 m is divisible by 6, say m 3 m = 6. Thus, (n + 1) 3 (n + 1) = (m 3 m) + 6(3m + 18m + 35) = 6 + 6(3m + 18m + 35), or (n + 1) 3 (n + 1) = 6(3m + 18m ). That is, (n + 1) 3 (n + 1) is divisible by 6 so by the Principle of Mathematical Induction, n 3 n is divisible by 6 for all n 0. Page 11

Some Review Problems for Exam 3: Solutions

Math 3355 Spring 017 Some Review Problems for Exam 3: Solutions I thought I d start by reviewing some counting formulas. Counting the Complement: Given a set U (the universe for the problem), if you want