MATH 3012 N Solutions to Review for Midterm 2

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1 MATH 301 N Solutions to Review for Midterm March 7, In how many ways can a n rectangular checkerboard be tiled using 1 and pieces? Find a closed formula. If t n is the number of ways to tile the n board, then t n = t n1 t n where t 1 = 1 and t = 3. Now we solve this recurrence: t n t n1 t n = 0 (A A t n = 0 (A (A 1t n = 0 Therefore, t n = c 1 n c ( 1 n. Now we solve for the constants using our initial conditions: 1 = c 1 c 3 = 4c 1 c Therefore c 1 = 3 and c = 1 3 so t n = 3 (n 1 3 ( 1n.. Let s(n 1 = s(n 3n with s(1 = 4. Find a closed formula for s(n. Observe s(n 1 s(n = 3n (A s n = 3n The general solution will be of the form s(n = c n d 1 n d n d 3. Put the particular solution into the recurrence to solve for d 1, d, and d 3. (A (d 1 n d n d 3 = d 1 (n n 1 d (n 1 d 3 d 1 n d n d 3 = d 1 n (d 1 d n (d 1 d d 3 Therefore d 1 = 3, d = 6, and d 3 =. So we have the solution s(n = c n 3n 6n. Now we can solve for c using the initial condition s(1 = 4. Thus 4 = s(1 = c 3 6 and c = 11 and the complete solution is s(n = 11 n 3n 6n. 3. A person deposits $1000 in an account that yields % interest compounded annually. After the interest has been calculated, he deposits another $100. (At the end of the first year, the account has $110. If the account has a(n dollars after n years, find a recurrence for a(n and solve the recurrence to give a closed formula for a(n. The recurrence is a(n 1 = a(n (.0a(n 100 = 1.0a(n 100 with a(0 = Now we can solve this recurrence. The closed form will look like a(n = c(1.0 n d.

2 First check the particular solution d: (A 1.0d = d 1.0d =.0d So let d = In the formula a(n = c(1.0 n 10000, solve for c using a(0 = We find c = = So the solution is ( 1000 a(n = (1.0 n A string of decimal digits is a valid codeword if it contains an even number of 0 digits. Let a(n be the number of valid codewords of length n. Find a recurrence for a(n and solve the recurrence. There are 10 n possible words. Of these, a(n have an even number of 0s and 10 n a(n have an odd number of 0s. Every codeword of length n 1 either ends in a 0 or a nonzero digit. If it ends in a nonzero digit, then the word on the first n digits form a valid codeword. If it ends in 0, then the first n digits must have an odd number of 0s. Therefore, we obtain the recurrence a(n 1 = a(n (10 n a(n = 8a(n 10 n. Now we solve this recurrence: (A 8a(n = 10 n. The solution has the form First check the particular solution: a(n = c8 n d10 n. (A 8d10 n = d10 n1 8d10 n = d10 n. Let d = 1. Now solve for c using the fact that there are valid codewords of length 1. a(n = c8 n 1 10n = 8c 5 c = 1 The complete solution is a(n = 1 8n 1 10n. 5. Find all solutions of the recurrence a n = 5a n1 6a n n 1 3n. The advancement operator equation is (A 5A 6a n = (A 3(A a n = 1 n 3n. So the homogenous solution will be of the form c 1 ( n c (3 n. Our first guess for a particular solution would be d 1 n d n d 3. However, n is already in the homogenous solution, so we should try d 1 n n d n d 3 for the particular solution. Now check the particular solution and solve for d 1, d, and d 3. (A 3(A (d 1 n n d n d 3 = (A 3(d 1 (n 1 n1 d (n 1 d 3 (d 1 n n d n d 3 = (A 3(d 1 n1 d n d d 3 = (d 1 n d (n 1 d d 3 (3d 1 n1 3d n 3d 3d 3 = d 1 n d n 3d d 3 Therefore d 1 = 1 4, d = 3, and d 3 = 4. The final solution is a n = c 1 ( n c (3 n 1 4 n(n 3 n Find the general solution for the advancement operator equation (A 1(A 5 k s(n = 0. [Yes. The k is supposed to be there this time.] The general solution is s(n = k c i n i 1 ( 5 n c k1 i n c k ( i n. i=1

3 7. How many positive integers not exceeding 100 are either odd or the square of an integer? The space is {1,,..., 100}. Consider the properties (1 an integer is odd and ( the integer is a square. There are 50 odd integers, 10 squares, and 5 which are both odd and a square. By inclusion-exclusion, the number of integers in {1,,..., 100} which are not odd and are not a square is = 45. Therefore, the number of integers in {1,,..., 100} which are either odd or a square is = How many ternary strings of length 8 have 6 consecutive 0s. The space is the set of all ternary strings of length 8. The 6 consecutive 0s could lie in positions 1-6, -7, or 3-8. For each i {1,, 3}, Property i states that there are 0s in the {i, i 1,..., i 5} entries. For each i {1,, 3}, N(i = 3. Then N(1, = N(, 3 = 3, N(1, 3 = 1, and N(1,, 3 = 1. Further, N( = 3 8. By inclusion-exclusion, the number of ternary strings which do not have 6 consecutive 0s is So the number of ternary strings of length 8 that have 6 consecutive 0s is There are 00 people at the airport. Each checked a single suitcase. At the luggage carousel, each person retrieves exactly one bag. However, exactly 5 people took the wrong bag. In how many different ways could the passengers have selected a bag off the carousel? First we must decide which 5 people pick up the wrong bag. There are ( 00 ways to select the 5 unlucky people. Now we just need a derangement of the suitcases for these 5 people because none of them received their own bag. So how many ways are there for these 5 unlucky passengers to grab a bag off the carousel. We will use inclusion exclusion for this. The space is the total number of ways for these 5 people to take a bag, with no restrictions. This is precisely the number of permutations of the 5 suitcases, which is 5!. Now we have 5 properties. For each i {1,, 3, 4, 5}, Property i says that person i got their own bag back. For any i {1,, 3, 4, 5}, N(i = 4! because person i must take their own bag and each of the other 4 people can take any one of the 4 remaining bags. More generally, N(S = (5 S!. By inclusion exclusion, the number of ways that these 5 people could select a suitcase that is not their own is ( ( ( ( ( ! 4! 3!! 1! This together with the choice of the 5 unlucky people gives a final count of ( ( ( ( ( ( ( ! 4! 3!! 1! A group of n students are assigned seats for each of two classes in the same classroom. How many ways can these seats be assigned if no student is assigned the same seat for both classes. Seat the students for the first class. There are ways to do this. Then for the second class, we need a derangement. Thus, the number of ways is n ( n ( 1 k (n k!. k k=0 See chapter 7 for the details on derangements. You should be able to give the details about where this expression came from. 11. Use inclusion exclusion to determine the number of integer solutions to the equation x 1 x x 3 = 5 such that x 1 8, 0 x 4 and 0 x 3 0. This is the same as asking for the number of solutions to x 1 x x 3 = 3 such that 0 x 1 6, 0 x 4 and 0 x 3 0. For a count by inclusion-exclusion, the space is the set of all non-negative integer solutions to x 1 x x 3 = 3. There are 3 properties: Property 1: x 1 7. Property : x 5.

4 Property 3: x 3 1. Therefore, N( = ( ( 3, N(1 = 3 7 (, N( = 3 5 (, N(3 = 3 1 (, N(1, = However N(1, 3 = N(, 3 = N(1,, 3 = 0 because you cannot have x 3 1 and x 1 7 or x 5 while adding to 3. By inclusion-exclusion, the answer to the original problem is ( ( ( ( ( In terms of partitions of integers, interpret the coefficient of x n in the following generating function: View the function as f(x = 1 x x 3 (1 x 7 (1 x 1. f(x = (1 x x (1 x 7 (1 x 1 = (1 x x 3 (1 x 7 x 14...(1 x 1 x This represents the number of ways to partition n into any number of 7 s, any number of 1 s and, if you like, you can use either a or a 3 (you cannot use both the and the 3, but you could use neither. 13. Use a generating function to determine the number of ways 15 identical stuffed animals can be given to 6 children so that each child receives at least one but no more than 3 stuffed animals. The generating function would be (x x x 3 6. It is more difficult to multiply this one out. But consider only the term x 15 in the product. This could come from x x x x 3 x 3 x 3 or x 1 x x 3 x 3 x 3 x 3. These are the only two possibilities. The first will come about in ( 6 3 ways. The second will happen in 6 5 ways. Thus, the number of ways to pass out the stuffed animals is ( Determine the coefficient of x 8 in each of the following generating functions: x (a f(x = (1 x 1 Therefore, the coefficient of x 8 is ( (b g(x = 1 1 3x 6x 1 x Observe x f(x = (1 x 1 1 = x (1 x 1 ( n = x (x n 0 0 ( n = n 0 x n 1 0 g(x = 1 1 3x 6 x 1 x = 1 1 3x 6 1 x x 1 x = 3 n x n 6 x n x Therefore the coefficient of x 8 is x n = 3 n x n 6x n x n1.

5 15. Determine the coefficient of x 301 in the exponential generating function (e x 4x(e x 0e 10x x. Find notice that x can be rewritten in a form that you recognize. In particular, x = e (ln x. Now, expand the generating function using summations. (e x 4x(e x 0e 10x x = (e x 4x(e x 0e 10x (ln x e = e x 0e 8x 4xe x 80xe 10x (ln x e ( 1 n x n 8 n x n x n = 0 4x 80x (10 n x n ( 1 n x n 0 8 n x n 4x n1 80(10 n x n1 = Therefore, the coefficient of x 301 is ( ! ! ! 80( ! (ln ! (ln n x n More likely, the question will ask for the coefficient of x ! since this is an exponential generating function. The answer here would be ( ( (ln Write a generating function for the number of ways in which postage of r cents can be pasted on an envelope using 3-cent, 4-cent, and 0-cent stamps. (The order of the stamps on the envelope does not matter. An ordinary generating function would be (1 x 3 x 6...(1 x 4 x 8...(1 x 0 x = 1 1 x x x Write a generating function for each of the following: You are making an Easter basket with n special edition Peeps. Each special edition Peep comes in its own one candy package. There are 4 different types of special edition Peeps: red velvet, lemon spice, party cake, and lemonade. Write a generating function for the number of ways to place n special edition Peeps in the basket if the basket can contain at most of these special edition Peeps. Note: If you have peeps, they could either be the same or different flavors. (Hint: Think about the corresponding sequence for different values of n. The basket can contain at most two of these special Peeps. So there should be a constant term, an x 1 term, and an x term in your generating function. The coefficient for x n is 0 when n 3. Now we must decide the coefficients for x 0, x 1, and x. There is 1 way to put no Peeps in the basket, so the constant term (or x 0 term is 1. There are 4 options for a single Peep, ( so the x 1 term should be 4x. If you have two Peeps, there are 4 ways to pick two of the same and 4 = 6 to have two different ones. So the x term should be 10x. All other terms should be 0 because you can have at most Peeps. So the generating function is 1 4x 10x. You are making an Easter basket and filling it with n pieces of candy. You can include any number of packages of 5 yellow chick Peeps, any number of packages of 5 blue bunny Peeps, and special edition peeps following the requirements in the previous problem. For the yellow chick Peeps, we will use the generating function (1 x 5 x = x. Similarly, for the blue bunny Peeps, you will need generating function 5 1 x. For the special 5 edition Peeps we will use the generating function we found in the previous problem. Therefore the generating function for the number of ways to fill a basket with n candies is x 5 1 x 5 (1 4x 10x. 18. Determine the closed form for the exponential generating function for each of the following sequences: (ln n x n

6 a n = 1 n1. The exponential generating function would be a n x n = 1 x n n 1 = x n (n 1! = 1 x x n1 (n 1! = ex 1. x b n = ( n. The exponential generating function would be x n b n = ( n xn = ( x n = e x. 1. Use a generating function to determine the number of strings of decimal digits of length n that have an odd number of 4s and at least one. We will need an exponential generating function because strings are ordered. For the digit 4, we use ex e x. For the digit, we use the generating function e x 1. For all other digits, we use e x because they can appear any number of times. So the generating function is ( e (e x 8 x e x (e x 1 = 1 (e10x e 8x e x e 7x ( = 1 (10x n (8x n (x n (7x n = 1 (10 n 8 n n 7 n x n 10 n 8 n n 7 n x n = Therefore the number of strings of length n is 10n 8 n n 7 n.

Exam 2 Solutions. x 1 x. x 4 The generating function for the problem is the fourth power of this, (1 x). 4

Exam 2 Solutions. x 1 x. x 4 The generating function for the problem is the fourth power of this, (1 x). 4 Math 5366 Fall 015 Exam Solutions 1. (0 points) Find the appropriate generating function (in closed form) for each of the following problems. Do not find the coefficient of x n. (a) In how many ways can