Some Review Problems for Exam 2: Solutions

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1 Math 5366 Fall 017 Some Review Problems for Exam : Solutions 1 Find the coefficient of x 15 in each of the following: 1 (a) (1 x) 6 Solution: 1 (1 x) = ( ) k + 5 x k 6 k ( ) ( ) 0 0 so the coefficient of x 15 is = = 15, (b) (1 + x ) 3 (1 x) 6 Solution: The coefficient of x 15 will be (c) 15!(1 + x ) 3 e 6x (1 + x ) 3 (1 x) 6 = (1 + 3x + 3x 4 + x 6 ) ( ) ( ) k + 5 x k k ( ) ( ) ( ) 14 = 56, Solution: The coefficient of x n in e 6x is 6n 615 We have 15!( n! 15! 13! or P (15, ) P (15, 4) P (15, 6) + 69 ), 11! 9! Find a closed form expression for the generating function for the sequence: (a) 5, 5, 5, 5, Solution: Answer: 5 + 5x + 5x + = 5 1 x (b) 5, 0, 5, 0, 5, 0, Solution: Here, the generating function is 5 + 5x + 5x 4 +, which is the same as in part (a), but with x replaced by x 5, so the answer is 1 x (c) 1, 6, 17, 34, (the k th term is 1 + k + 3k ) Solution: We could break up the problem to finding 3 different generating functions: x k + kx k + 3k x k But it is usually easier to use the trick

2 ( ) m + k that x k 1 = k (1 x) To use the trick, try to write m+1 1+n+3n as a sum of binomial coefficients of the appropriate type This is another version of an undetermined coefficients problem: Write ( ) ( ) n + 1 n n + 3n = A + B + C 1 When n = 0 we get A + B + C = 1, when n = 1, A + B + 3C = 6, and when n =, A + 3B + 6C = 17 This system has solution A =, B = 7, C = 6 so (1 + k + 3k )x k = ( ( ) k ( k = 1 x 7 (1 x) + 6 (1 x) 3 )) x k 3 Use the answer to part (c) above to find a formula for (1 + n + 3n ) Solution: We need the coefficient of x n in 1 A(x), where A(x) is the generating 1 x function for the sequence in part (c) That is, we need the coefficient of x n in or (1 x) 7 (1 x) (1 x), 4 ( ) ( ) ( ) n + 1 n + n = n3 + 5n + 5n + 4 Find the generating function for each of the following problems Write your answer as simply as possible (that is, 1 + x + x 3 + x 4 + should be written 1 1 x+x x or ) 1 x 1 x Note: Just find the generating function (ordinary or exponential,) don t solve for x n (a) In how many ways can n balls be put into 5 boxes if there is at least one ball in the first box and at most 3 in the second? Solution: x 1 x (1 + x + x + x ) 1 x 1 x 1 x = x(1 + x + x + x 3 ), or (1 x) 4 x x 5 (1 x) 5 Page

3 (b) How many sequences made up of a, b, c, d, e of length n are there if there is at least one a and at most 3 b s? Solution: (e x 1) ) ) (1 + x + x! + x3 e x e x e x = (1 + x + x 3!! + x3 (e 4x e 3x ) 3! (c) How many ways can n balls be put into 5 boxes if the first box has an even number of balls and the second, an odd number? Solution: The generating function for an even number of balls in a box is 1 + x + x 4 + = 1 For an odd number of balls, you just multiply this by 1 x x The generating function we seek is x (1 x ) (1 x) 3 (d) How many sequences made up of a, b, c, d, e of length n are there if there must be an even number of a s and an odd number of b s? Solution: Here, the generating functions for even/odd are ex + e x The generating function we seek is and ex e x e x + e x e x e x e x e x e x = e5x e x 4 (e) How many ways can n balls be put into 5 boxes if no box has exactly balls? Solution: For one box it would be 1 + x + x 3 + x 4 + = 1 1 x x = 1 x + x 3 Probably the first formula is better Our generating function is 1 x ( ) x x Page 3

4 (f) How many sequences made up of a, b, c, d, e of length n are there if you can t have exactly of any letter? Solution: Here there really only is one answer: ) 5 (e x x 5 Find the solutions to the problems in question 1 (a) In how many ways can n balls be put into 5 boxes if there is at least one ball in the first box and at most 3 in the second? Solution: We seek the coefficient of x n in x x5 This will be (1 x) 5 ( ) ( ) ( ) ( ) n n n + 3 n 1 = I( will note that this formula is correct if n > 0 (rather than just n > 5) because n ) k is defined to be 0 k > n 0 (b) How many sequences made up of a, b, c, d, e of length n are there if there is at least one a and at most 3 b s? Solution: Here we are stuck ( with a rather complicated ) expression We need n! times the coefficient of x n in 1 + x + x + x3 (e 4x e 3x ) We get! 3! 4 n 3 n +n(4 n 1 3 n 1 )+ n(n 1) (4 n 3 n )+ I don t know of any good way to simplify this n(n 1)(n ) (4 n 3 3 n 3 ) 6 (c) How many ways can n balls be put into 5 boxes if the first box has an even number of balls and the second, an odd number? Solution: We need the coefficient of x n x in This is just a hard (1 x ) (1 x) 3 problem If we were to use partial fractions, we would have x (1 x ) (1 x) = x 3 (1 + x) (1 x) 5 = 1 ( x 4 (1 x) + 4 (1 x) (1 x) 3 ) x (1 + x) Page 4

5 Our coefficient would be ( ( ) ( ) ) 1 n + n (n + 1) ( 1) n (n + 1)( 1) n 64 4 (d) How many sequences made up of a, b, c, d, e of length n are there if there must be an even number of a s and an odd number of b s? Solution: This problem is much easier as we want n! times the coefficient of x n in e5x e x, or 5n (e) How many ways can n balls be put into 5 boxes if no box has exactly balls? Solution: We have ( ) x 1 x = (1 x) 5x 5 (1 x) + 10x4 4 (1 x) 10x6 3 (1 x) + 5x8 1 x x10 so if n > 10 we get ( ) ( ) ( ) n + 4 n + 1 n (n 5) + 5, 4 3 with various other answers for smaller n (f) How many sequences made up of a, b, c, d, e of length n are there if you can t have exactly of any letter? Solution: This time, the two cases are fairly similar in complexity We have ) 5 (e x x = e 5x 5x e 4x + 10x 4 e 3x 10x 6 e x + 5x 8 e x x 10 so for n > 10 the answer is 5 n 5n(n 1)4 n + 10P (n, 4)3 n 4 10P (n, 6) n 6 5P (n, 8) Technically, we only need worry about n = 10 as an exceptional case here, since P (n, k) = 0 if n < k When n = 10, we must subtract 10! from the expression above (with n replaced by 10 everywhere) Page 5

6 6 Use generating functions to find the number of ways to put n balls into 3 boxes if the first box can hold at most 5 balls and the second can hold at most 10 Your answer will vary depending on the size of n For large n, simplify your answer as much as possible Can you prove your answer (for large n) by an easy counting argument? Solution: The generating function is or (1 + x + x + x 3 + x 4 + x 5 )(1 + x + + x 10 )(1 + x + x + ), (1 x 6 )(1 x 11 ) (1 x) 3 = 1 x6 x 11 + x 17 (1 x) 3 The coefficient of x n in this expression, for large n is ( ) ( ) ( ) ( ) n + n 4 n 9 n 15 + Expanding this out and simplifying, you get a constant, 66 The reason: for large n, there are 6 choices for how many balls to put in the first box and 11 for the second What s left goes in the third box So if n 17 we get 6 11 = 66 possible combinations 7 Find recurrences, with initial conditions, for each of the following problems (a) How many n-digit ternary sequences do not contain consecutive 0 s? Solution: This is kind of easy after the last few questions The reason: This problem (both parts) came from a past exam If the first position is a 1 or a, the remaining n 1 positions just have to avoid consecutive 0 s This can happen in a n 1 ways If the first position is a 0 then the second position must contain 1 or, and the remaining n positions just need to avoid consecutive 0 s The relation is a n = a n 1 + a n For initial conditions, the empty sequence does not contain consecutive 0 s, nor does any sequence of length 1 so we have a 0 = 1, a 1 = 3 If you worry about whether the empty sequence has some property, you can find a 1 and a instead In this case there is only one bad string of length so a = 9 1 = 8 (b) How many n-digit ternary sequences contain consecutive 0 s? Solution: Usually, the recurrence for this kind of problem is the same as for without, except that there will be an inhomogeneous term The count goes like this: If the first position is a 1 or a, then we need to pick up consecutive 0 s Page 6

7 among the remaining n 1 positions for a count of a n 1 If the first position is a 0 but the second is 1 or, we still need consecutive 0 s, and they must come from the remaining n positions for a count of a n But here, we have another case: If the first two positions are 0 s, we have our consecutive 0 s and the remaining n positions are arbitrary The recurrence is a n = a n 1 + a n + 3 n Here, the initial conditions are a 0 = 0, a 1 = 0, or if you prefer, a 1 = 0, a = 1 (c) How many n-digit quaternary sequences do not contain any of 00, 10, 10, 11? Solution: First, there is a typo: It should have been 00, 01, 10, 11 that we are avoiding If the first position is a or a 3, then we need only avoid 00, 01, 10, 11 in the remaining n 1 positions This leads to a count of a n 1 If we start with a 0 or a 1, then the next position must be a or a 3, and the remaining n positions must still avoid 00, 01, 10, 11 Since there are two possibilities for each of the first two positions, this case has a count of 4a n Our recurrence is a n = a n 1 + 4a n, with initial conditions a 0 = 1, a 1 = 4 (d) How many n-digit quaternary sequences do not have any 0 s or 1 s to the right of any 3? Solution: If we start with 0, 1 or, then we need only avoid 3 s to the right of 0 s or 1 s the rest of the way for a count of 3a n 1 But if the first position is a 3, then no subsequent position can be 0 or 1, leaving only two choices ( or 3) for each of the remaining n 1 positions, for a count of n 1 Our recurrence is a n = 3a n 1 + n 1 with initial condition a 0 = 1 8 Solve the recurrence in question 7(d) Solution: For the practice, I will use the iterative approach We have a 0 = 1, a 1 = 3 + 1, a = , a 3 = , This might be enough to guess that a n will have two pieces, 3 n and a geometric piece 3 n n + 3 n n 1 If we multiply this second piece by 3 - it becomes 3 n n Thus, it appears that a n = 3 n n To check, we show that this satisfies the recurrence If a n 1 = 3 n 1 n 1 then a n = 3a n 1 n 1 = 3( 3 n 1 n 1 ) + n 1 = 3 n 3 n 1 + n 1 = 3 n n, as desired Page 7

8 9 I once put the following problem on Exam, but the class found it fairly hard (under exam conditions) Suppose we form a sequence out of the numbers 0, 1,, 3, 4, 5 (a) Find a recurrence for a n, the number of such sequences of length n which don t contain consecutive numbers not divisible by 3 That is, sequences like are ok but 1343 is not because 4 and are consecutive numbers not divisible by 3 Solution: o and 3 are divisible by 3, and 1,, 4, 5 are not If the first position is 0 or 3, we just need to avoid consecutive numbers not divisible by 3 in the remaining n 1 positions, contributing a n 1 If the first digit is 1,, 4, or 5, then the next digit must be 0 or 3, and we need to avoid numbers not divisible by 3 in the remaining n positions This gives a count of 4 a n for this case The recurrence is a n = a n 1 + 8a n (b) Find the initial conditions for a n Solution: As usual, sequence of length 0 or 1 are to short to cause problems so a 0 = 1, a 1 = 6 If you want a 1 and a in stead, we must do a little more work There are 36 sequences of length, and 4 4 = 16 of them are bad so a = 0 (c) Solve the recurrence There should be fractions but no radicals Solution: This was the point of the problem: I was proud of myself for coming up with a fairly natural recurrence that had a nice solution The characteristic polynomial is x x 8 = (x + )(x 4) The general solution is a n = A( ) n + B4 n The initial conditions tell us A + B = 1, A + 4B = 6 so B = 4, A = 1 Our solution is a 3 3 n = 4n+1 ( ) n 3 (d) Find a recurrence for the number of such sequences of length n which DO contain consecutive numbers not divisible by 3 Now is not ok but 1343 is not Another typo here it was supposed to say that 143 IS OK Solution: If we start 0 or 3, we need consecutive numbers not divisible by 3 in the remaining n 1 positions If we start with 1,, 4, 5 but follow with 0 or 3, then we need to fulfill the requirements in the remaining n positions Finally, if the first and second positions contain any of 1,, 4, 5, then the remaining n positions are arbitrary The recurrence is a n = a n 1 + 8a n n The initial conditions are a 0 = 0, a 1 = 0 Page 8

9 10 Solve the following recurrence relations (a) a n = 3a n 1 + 4a n 1a n 3, a 0 = 1, a 1 = 4, a = 14 Solution: The characteristic polynomial is x 3 3x 4x+1 = (x )(x+)(x 3) The general solution is a n = A n + B( ) n + C3 n The initial conditions give A + B + C = 1, A B + 3C = 4, 4A + 4B + 9C = 14, which has solution C =, A = 1, B = 0 The solution is a n = 3 n n (b) a n = 3a n 1 + 4a n 1a n n, a 0 = 1, a 1 =, a = 6 Solution: We try for particular solution a n = P 4 n, which gives P 4 n = 3P 4 n 1 + 4P 4 n 1P 4 n n Dividing by 4 n, we have 16P = 1P + 4P 3P + 3, or P = 1 The general solution is a n = A n + B( ) n + C3 n + 4 n, and with the initial conditions, A + B + C = 0, A B + 3C = 6, 4A + 4B + 9C = 10, and we have A = 1, B = 1, C = so a n = 4 n 3 n + n + ( ) n (c) a n = 7a n 1 16a n + 1a n 3, a 0 = 1, a 1 =, a = 6 Solution: The characteristic polynomial is x 3 7x +16x 1 = (x ) (x 3) The general solution is a n = (A + Bn) n + C3 n, and the initial conditions tell us A + C = 1, A + B + 3C =, 4A + 8B + 9C = 6 The solution to these is A = 1, B = 1, C = so a n = 3 n (n + 1) n (d) a n = 7a n 1 16a n + 1a n 3 n, a 0 = 4, a 1 = 7, a = 1 Solution: Since is a double root of the characteristic polynomial we can t use a n = P n or even np n We try a n = n P n We have a mess: n P n = 7(n 1) P n 1 16(n ) P n + 1(n 3) P n 3 n Dividing by n 1, P n = 7P (n 1) P 8P (n ) + 3P (n 3), or P (n 7n + 14n 7 + 8n 3n + 3 3n + 18n 7) = This simplifies to P =, or P = 1 The fact that the n terms and n terms cancelled tells us that our guess was right The general solution is a n = (A+Bn) n +C3 n + n, and the initial conditions give A+C = 4, A+B+3C = 5, 4A + 8B + 9C = 5, leading to A = 3, B =, C = 1 The solution to the recurrence is a n = 3 n + (n n + 3) n Page 9

10 (e) a n = a n 1 a n, a 0 = 1, a 1 = 3, USING generating functions Solution: We let A(x) = recurrence: A(x) = a 0 + a 1 x + a n x n We pull off the first two terms and use the n=0 a n x n = a 0 + a 1 x + (a n 1 a n )x n We break the sum into two pieces a n 1 x n, a n x n The second of these is a 0 x a 1 x 3 a x 4 + = x A(x) For the first, we do something similar Factoring an x out we have x(a 1 x + a x + ) = x(a(x) a 0 ) Putting the pieces together, A(x) = a 0 + a 1 x + x(a(x) a 0 ) x A(x) We solve this equation for A(x) and we have (x x + 1)A(x) = a 0 + x(a 1 a 0 ) Plugging in the initial conditions, 1 + x A(x) = x x + 1 = 1 + x (1 x) ( ) ( ) n + 1 n a n is the coefficient of x n in A(x), or + = n (f) a n = 6a n 1 9a n + 3 n, a 0 = 1, a 1 = 0, with and without generating functions Solution: Without generating functions: The characteristic polynomial for the homogeneous problem is x 6x + 9 = (x 3), so the particular solution must have the form a n = P n 3 n Plugging in, P n 3 n = 6P (n 1) 3 n 1 9P (n ) 3 n + 3 n We divide by 3 n giving P n = P (n 1) P (n ) + So P (n n + 4n + n 4n + 4) = or P = The general solution is a n = (A + Bn)3 n + 3 n, and the initial conditions tell us A = 1, B = so a n = (n n + 1)3 n With generating functions: A(x) = a n x n = a 0 +a 1 x+ a n x n = a 0 +a 1 x+ (6a n 1 9a n + 3 n )x n n=0 Page 10

11 we have 6a n 1 x n = 6x a n 1 x n 1 = 6x(A(x) a 0 ), 9a n x n = 9x a n x n = 9x A(x) 3 n x n = 18x 3 n x n = 18x 1 3x and so A(x) = a 0 +a 1 x+6x(a(x) a 0 ) 9x A(x)+ 18x 1 3x, or (1 6x+9x )A(x) = a 0 + a 1 x 6xa x 1 3x = 1 6x + 18x 1 3x Thus, A(x) = and the coefficient of x n is ( ) ( ) n + 1 n 3 n 6 3 n as before ( n 1 6x (1 3x) + 18x (1 3x) 3 ) 3 n = 3 n (n+1 n+n n) = (n n+1)3 n, (g) a n = na n 1 + n n!, a 1 = 1, with and without generating functions For the generating function approach, use exponential generating functions Solution: One way to avoid generating functions is to make a change of variables to get constant coefficients If we write a n = n! b n then the recurrence becomes n! b n = n((n 1)! b n 1 ) + n n! = n!(b n 1 + n ) or b n = b n 1 + n If a 0 = 1 and b 0 = 1 as well Iteration tells us b 1 = +, b = + +, etc so we must have b n = (n + 1) n giving a n = (n + 1)! n Using exponential generating functions, we let A(x) = the first term and use the recurrence: A(x) = a 0 + n=0 x n a n n! = 1 + (na n 1 + n n!) xn n! x n a n We pull off n! The point of exponential generating functions is that they help the factor of n Page 11

12 go away That is, n n! = 1 (n 1)! Thus, we have A(x) = 1 + x n a n 1 (n 1)! + n x n = 1 + x x n 1 a n 1 (n 1)! + x 1 x, or A(x) = 1 + xa(x) + x 1 x = xa(x) x, or A(x) = 1 (1 x) The coefficient of x n in this expression is (n+1) n, but we need n! times this number giving a n = n!(n + 1) n = (n + 1)! n 11 Solve each of the following recurrence relations (a) a n = a n 1 + 4a n + 4a n a 0, a 0 = 1, a 1 = 10 Solution: This can probably be done with generating functions but the following is easier If n 3 then a n 1 = a n + 4a n a 0 and if we subtract this from the original we have a n a n 1 = a n 1 + 3a n, or a n = a n 1 + 3a n Now we have a second order recurrence, but only from n = 3 on It has characteristic polynomial x x 3 = (x 3)(x + 1) The general solution has the form a n = A3 n + B( 1) n We can t plug a 0 and a 1 into this, we need a 1 and a instead We use the original recurrence to tell us a = a 1 + 4a 0 = 14 Now 3A B = 10, 9A + B = 14 so A =, B = 4 Our solution is a 0 = 1, a n = 3 n 4( 1) n for n 1 As a check, when n = 3 we have a 3 = a + 4a 1 + 4a 0 = = 58 and 3 3 4( 1) 3 = = 58 (b) a n = a n 1 + a n + 4a n n 1 a 0, a 0 = 1 Solution: This can be done fairly easily with generating functions because a n is almost a convolution (of the a s and n ) We have A(x) = a n x n = a 0 + a n x n = a 0 + (a n 1 +a n +4a n n 1 a 0 )x n n=0 If we change the index on this last sum we have (a n +a n 1 +4a n + + n a 0 )x n+1 = x (a n +a n 1 +4a n + + n a 0 )x n, n=0 which is a convolution (again, of a 0 + a 1 x + a x + and 1 + x + x + ) 1 This gives A(x) = a 0 + xa(x) 1 x = 1 + A(x) x Solving for A(x), 1 x ( 1 x ) 1 3x A(x) = 1, or A(x) = 1 1 x 1 x n=0 Page 1

13 Thus, A(x) = 1 x 1 3x, and a n is the coefficient of x n in this That coefficient is 1 when n = 0 and 3 n 3 n 1 = 3 n 1 when n 1 So for n 1, a n = 3 n 1 Easier is to use the same trick as in part (a) We can t just subtract a n 1, in this case When n we can write a n 1 = a n +a n n a 0 If we multiply this by and subtract from the original recurrence we have a n a n 1 = a n 1, or a n = 3a n 1 Noting that a 1 = 1, ti follows that a n = 3 n 1 for all n 1 Finally, I will mention that on this problem, it is easy to guess the answer We have a 0 = 1, a 1 = a 0 = 1, a = a 1 + a 0 = 3, a 3 = a + a 1 + 4a 0 = 9 If we guess a 0 = 1, a n = 3 n 1 for n 1 then a n = a n 1 + a n + + n a 1 + n 1 a 0 = 3 n + 3 n n n + n 1 All but the last term are part of a geometric series with common ratio 3 so a n = 3n 3 n n 1 = 3 n 1 n 1 + n 1 = 3 n 1, verifying that the formula is correct Page 13

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