COMPSCI 575/MATH 513: Combinatorics and Graph Theory

Solutions to Practice Second Midterm Exam, Fall 2016

David Mix Barrington

14 November 2016

Directions:

Answer the problems on the exam pages.
There are four problems (some with multiple parts) for 100 total points. Estimated scale is A = 93, C = 63.
If you need extra space use the back of a page.
No books, notes, calculators, or collaboration.
Numerical answers may be given as expressions involving arithmetic operations (including exponentiation and factorial) or the functions P(n, k) and C(n, k). In general a summation is a complete answer to a counting problem, but a recurrence or a generating function is not.

  Q1: 20 points
  Q2: 40 points
  Q3: 20 points
  Q4: 20 points
Total: 100 points

Question text is in black, solutions in blue.

Correction in green made 16 November 2016.

Correction in orange made 17 November 2016.

Question 1 (20): Identify each of these statements as true or false. No explanation is needed or wanted, and there is no partial credit or penalty for guessing (2 points each).
- (a) There are P(n, k)² ways to place k identical noncapturing rooks on an n by n chessboard.
  FALSE. The correct number is given in part (c).
- (b) There are P(n, k)C(n, k)² ways to place k identical noncapturing rooks on an n by n chessboard.
  FALSE. I meant for this to be true, because I meant to write "P(n, k)C(n, k)" without the C(n, k) being squared. That would have been correct since we could pick a set of k rows in C(n, k) ways, then place a rook in the first selected row in k ways, the next in k-1, and so on.
- (c) There are k!C(n, k)² ways to place k identical noncapturing rooks on an n by n chessboard.
  TRUE. If we choose a set of rows and a set of columns, there are k! ways to place noncapturing rooks on those rows and columns.
- (d) The number of ways to distribute r identical treats to r identical dogs is C(n+r-1, n-1).
  FALSE. This is the number of ways to distribute r identical treats to n distinct dogs.
- (e) There are exactly fifteen ways to distribute seven identical treats to seven identical dogs, where we don't insist that each dog gets a treat.
  TRUE. The partition number P(7) is 15, for the partitions 7, 61, 52, 511, 43, 421, 4111, 331, 322, 3211, 31111, 2221, 22111, 211111, and 1111111.
- (f) There is no self-conjugate partition of 7.
  FALSE. 4111 is self-conjugate (it is the only one).
- (g) If A, B, and C are any three sets, |A ∩ B ∩ C| = |A| + |B| + |C| - |A ∪ B| - |A ∪ C| - |B ∪ C| + |A ∪ B ∪ C|.
  FALSE. This formula fails if A = B = C = {x}. The LHS is 1 but the RHS is 3 - 6 + 3 = 0.
- (h) Any recurrence of the form a_n = ca_n-1 + d, where c and d are integers, has a solution of the form a_n = dn + e, where e is an integer depending on a₀.
  FALSE. If c = 1 this works, but if c ≠ 1 it won't in general. For example, if c = 2 and d = 1, a_n = 2ⁿa₀ + 2ⁿ - 1, which is not linear unless a₀ = 0, and does not have slope d if a₀ = 0.
- (i) If f(x) and g(x) are exponential generating functions for the sequences a₀, a₁,... and b₀, b₁,... respectively, then f(x)g(x) is the exponential generating function for c₀, c₁,..., where for each n c_n is the sum for i from 0 to n of a_ib_n-i.
  FALSE. This is the rule for ordinary generating functions, but for example if a_n = b_n = 1 for all n, f(x) = g(x) = e^x, so f(x)g(x) = e^2x which is the EGF for 1, 2, 4, 8,..., not for 1, 2, 3, 4,...
- (j) For any n, the sum for k from 0 to n of (-1)^k2^n-kC(n, k) is 1.
  TRUE. This is the expansion by the Binomial Theorem of (-1+2)ⁿ.
Question 2 (40): In this problem we have a set of eleven dog treats {t₁,..., t₁₁} and a set of four dogs {Arly, Biscuit, Cardie, Duncan}. We will investigate the distribution of these treats to the dogs.
- (a, 10) In how many ways can the eleven treats (considered here to be identical) be distributed to the four dogs, if each dog must get at least two treats? You many use any method, but explain how you got your answer.
  Give each dog two treats, then choose a multiset of size 3 from the four dogs to distribute the other three. By the stars and bars argument, there are C(3+4-1, 4-1) = C(6, 3) = 20 such multisets.
- (b, 10) Suppose now we consider the treats to be distinct, but still require that each dog get at least two treats. Give an exponential generating function for the number of ways to distribute r distinct treats to four distinct dogs, and indicate how to obtain the answer for eleven treats from this function. (You need not actually determine the answer.)
  The EGF for the number of ways to distribute at least two distinct treats to one dog is x²/2! + x³/3! + ... = e^x - 1 - x. So the EGF for four dogs is (e^x - 1 - x)⁴. We would need to multiply out this function, determine its x¹¹ coefficient, and multiply that coefficient by 11!.
- (c, 10) Let X be the set of all distributions of eleven identical treats to the four distinct dogs. Let A be the subset of X consisting of distributions that give Arly zero or one treats, and let B, C, and D be the analogous sets for Biscuit, Cardie, and Duncan. Determine the sizes of X, A, B, C, and D, and show how the answer for part (a) may be determined from these by Inclusion/Exclusion.
  The size of X is C(11+4-1, 4-1) = C(14, 3). The size of A is the number of ways to give either ten or eleven treats to the other three dogs, which is C(10+3-1, 3-1) + C(11+3-1, 3-1) = C(12, 2) + C(13, 2), and of course B, C, and D have the same size. To use I/E, we need |X| - |A| - |B| - |C| - |D| + (C(4, 2) terms like |A∩B|) - (C(4, 3) terms like |A∩B∩C| + (last term which is zero because no distribution in X is in all four sets). |A∩B| is the number of ways to give zero or one treat to each of Arly and Biscuit, and the rest to the other two dogs, which is C(11+2-1, 2-1) + 2C(10+2-1, 2-1) + C(9+2-1, 2-1). |A∩B∩C| is just the number of ways to give zero or one treat to each of the first three dogs, since for each of these there is only one way to give Duncan all the other treats.
- (d, 10) Give an ordinary generating function for the number of ways to give r identical treats to the four dogs, with each dog getting at least two treats. Express your generating function as a ratio of polynomials. Indicate how the answer for part (a) may be obtained from this function (you need not actually do it).
  The GF is (x² + x³ + ...)⁴, which we can express as (x²/(1-x))⁴ = x⁸/(1-x)⁴. We want the x¹¹ coefficient of this GF, which is the x³ coefficient of 1/(1-x)⁴.
Question 3 (20): These questions involve the recurrence a_n = 3a_n-2 + 2a_n-3, with initial conditions a₀ = 2, a₁ = 0, and a₂ = 7.
- (a, 5) Determine the value of a₇, by any method.
  a₃ = 3a₁ + 2a₀ = 3(0) + 2(2) = 4.
  a₄ = 3a₂ + 2a₁ = 3(7) + 2(0) = 21.
  a₅ = 3(4) + 2(7) = 26.
  a₆ = 3(21) + 2(4) = 71.
  a₇ = 3(26) + 2(21) = 120.
- (b, 5) Give the characteristic polynomial of this recurrence, and factor it as a product of linear polynomials.
  The polynomial is x³ - 3x - 2 = 0.
  One of the roots is 2 since 2³ - 3(2) - 2 = 0. (We might guess 2 as a root because the solutions in part (a) start to look like they are around doubling each time.) Dividing the polynomial by x - 2 gives us x² + 2x + 1, so the other two roots are each -1.
- (c, 10) Determine the general solution for this recurrence, and the specific solution for these initial conditions.
  The general solution is A2ⁿ + (Bn + C)(-1)ⁿ. The n = 0 case tells us that A + C = 2. The n = 1 case tells us that 2A - (B + C) = 0, and the n = 2 case tells us that 4A + (2B + C) = 7. Subtracting the first equation from the third says that 3A + 2B = 5, and adding the first two tells us that 3A - B = 2. This solves to A = B = C = 1, so the solution is a_n = 2ⁿ +(-1)ⁿ(n + 1) which fits the results in part (a).
Question 4 (20): Define the function g(n) to be the sum, for i from 0 to n, of C(n - i, i). Define the function f(n) by the recurrence f(n) = f(n-1) + f(n-2), with initial conditions f(0) = f(1) = 1. (This is Tucker's form of the Fibonacci seqeunce.) Prove, by any method, that f and g are the same function.
We can just prove that g(n) satisfies the recurrence and the initial conditions. First, g(0) is C(0, 0) which is 1. Next, g(1) is C(1, 0) + C(0, 1) = 1 + 0 = 1 as well.
Finally, g(n-1) + g(n-2) is the sum for i from 0 to n-1 of C(n-1-i, i) plus the sum for i from 0 to n-2 of C(n-2-i, i). We can write the second sum as the sum for i from 1 to n-1 of (C(n-1-i, i-1). Note that this sum is the same if we make it for i from 0 to n-1, since C(n-1-0, 0-1) = 0.
The total is the sum for i from 0 to n-1 of [C(n-1-i, i) + C(n-1-i,i-1)]. By Pascal's Identity, the expression in brackets is C(n-i, i), so the sum is the desired one for g(n).

Last modified 17 November 2016