COMPSCI 575/MATH 513: Combinatorics and Graph Theory

Solutions to Second Midterm Exam, Fall 2016

David Mix Barrington

17 November 2016

Directions:

• Answer the problems on the exam pages.
• There are four problems (some with multiple parts) for 100 total points. Actual scale was A = 90, C = 54.
• 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: 25 points
  Q4: 15 points
Total: 100 points

Question text is in black, solutions in blue.

• 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) The rook polynomial for a 2 × 3 rectangle is 1 + 6x + 9x².

    FALSE. The 1 and 6x are correct but it should be 6x².

  • (b) The rook polynomial for the board made by five squares, arranged in a plus sign, is 1 + 5x + 4x² + x³.

    FALSE. The x³ term should be deleted.

  • (c) If |A| = 5 and |B| is 4, then the number of onto functions from A to B is exactly 4×C(5, 2)×3!.

    TRUE. We choose which element of B is hit twice, then which two elements of A hit it, then how the other three elements of A go 1-1 to the other three elements of B.

  • (d) The number of partitions of 7 into three non-empty parts is equal to the number of partitions of 7 into parts whose size is at most three.

    FALSE. If we change the first condition to "at most three parts" it is correct. We could also change the second condition to "with largest part of size exactly three".

  • (e) The recurrence a_n = a_n-1 + a_n-3 + n³ - 6 has a unique solution with the initial conditions a₀ = 0 and a₁ = 1.

    FALSE. The value x₂ could be anything. Since this is a third-order recurrence, we would need three initial conditions to determine it uniquely, but we are given only two.

  • (f) If f(x) = 1 + a₁x + a₂x² + a₃x³ + ... is any power series with integer coefficients, then there exists a power series g(x) with integer coefficients such that g(x)(1 + x + x² = f(x).

    TRUE. We can find g(x) in principle by long division, since the constant term of the dividend is 1.

  • (g) The number of ways to distribute r distinct treats to n identical dogs is n^r.

    FALSE. That is the number of ways to distribute r distict treats to r distinct dogs.

  • (h) Consider paths in a Manhattan-like grid where edges go only in the positive x or positive y direction. The number of paths from (0, 0) to (n, n) is equal to the sum, for i from 1 to n, of C(n, i)C(n, n-i).

    FALSE. The sum should be from 0 to n.

  • (i) For any n, the number of ternary sequennces of length n with at least one occurrence of each letter is 3ⁿ - 3×2ⁿ + 3×1ⁿ - 1.

    FALSE. The last "-1" term should be "-1×0ⁿ", i.e., 0.

  • (j) The words BASINGSTOKE and BELCHERTOWN each have exactly P(11, 2) = 11!/9! anagrams. (An anagram of a word is any word (whether a valid English word or not) containing the same multiset of letters.)

    FALSE. Both have 11!/2! anagrams, since the 11! permutations of each count each word twice. Exchanging the two S's in BASINGSTOKE or the two E's in BELCHERTOWN leads to the same anagram.

• Question 2 (40): In this problem we distribute a set of 13 treats {t₁,..., t₁₃} to the set of dogs {c, d, s} (Cardie, Duncan, and Scout). We will insist that Cardie and Duncan each get at least three treats, and that Scout gets an even number of treats (possibly 0).
  • (a, 10) Determine, by any method, the number of ways to distribute 13 identical treats to the three distinct dogs, with the requirements above.

    It is perhaps easiest to begin by giving three treats each to Cardie and Duncan, then considering the ways to distribute the other seven treats. Scout could get 0, 2, 4, or 6 of these. If she gets 0, there are C(7+2-1, 2-1) = 8 ways to divide the other 7 treats among the other two dogs. If Scout gets 2, there are C(6+2-1, 2-1) = 6 ways to divide the others. Similarly there are 4 ways if Scout gets 4 and 2 ways if Scout gets 6, for 8+6+4+2 = 20 ways in all.

  • (b, 10) Find an ordinary generating function whose x^r coefficient is the number of ways to distribute r identical treats to the three distinct dogs, while following the requirements stated above. Determine the x¹³ coefficient of this generating function.

    For Cardie and Duncan, the GF for the ways to give them three or more treats is x³ + x⁴ + x⁵ + ... = x³/(1-x). For Scout, the GF for the number of ways to give her an even number of treats is 1 + x² + x⁴ + ... = 1/(1-x²). (Someone wrote this as (1/2)[1/(1-x) + 1/(1+x)], which is also correct.)

    Writing (1-x²) as (1-x)(1+x), the total GF evaluates to x⁶ × 1/(1-x)³ × 1/(1+x). To get the x¹³ coefficient of this, we need the x⁷ coefficient of f(x)g(x), where f(x) = 1/(1-x)³ = ∑C(r+1, 2)x^r, and g(x) = 1/(1+x) = ∑(-1)^rx^r. So we have the sum for i from 0 to 7 of C(i+1, 2)(-1)^7-i, or -1 + 3 - 6 + 10 - 15 + 21 - 28 + 36 = 2 + 4 + 6 + 8 = 20.

  • (c, 10) Find an exponential generating function for the number of ways to distribute r distinct treats to the three distinct dogs, meeting the requirements above. Express this function using exponential functions and polynomial. You need not evaluate its x¹³ coefficient, but give a brief idea of how you would do so.

    The EGF for Cardie and Duncan is each x³/3! + x⁴/4! + x⁵/5! + ... = e^x - 1 - x - x²/2. The EGF for Scout is 1 + x²/2! + x⁴/4! + ... = (e^x + e^-x)/2. We multiply these three expressions together (getting a sum of 32 terms), evaluate the x¹³ coefficients of each, and add the answers. This coefficent, multiplied by 13!, is the number of ways to give 13 distinct treats to the three dogs, following the conditions.

  • (d, 10) Express the solution to part (a) using Inclusion/Exclusion, where N is the total number of ways to distribute 13 identical treats to three distinct dogs, and A₁, A₂, and A₃ are the sets of distributions that violate each of the three conditions abobve, respectively. Find the sizes of N, A₁, A₂, and A₃ -- you need not find the sizes of the various intersections but you should express your final answer in terms of them.

    The total number N is C(13+3-1, 3-1) = C(15, 2) = 105 by a stars and bars argument. To determine the size of A₁ and A₂, we break into three cases by whether the dog in question violates the condition by getting 0, 1, or 2 treats. In each case the remaining treats must be distributed between the other two dogs, in 14, 13, or 12 ways respectively, so these two sets each have size 14+13+12 = 39. To get the size of A₃, we divide into cases for Scout getting 1, 3, 5, 7, 9, 11, or 13 treats. In each case the others are distributed between Cardie and Duncan, in respectively 13, 11, 9, 7, 5, 3, and 1 ways, for 49 ways in total.

    So the correct number of ways to follow the rules is 105 - 39 - 39 - 49 + |A₁ ∩ A₂| + |A₁ ∩ A_{3/sub>| + |A2 ∩ A3| - |A1 ∩ A2 ∩ A3|.}

    You were not required to go further than this, but the first two two-way intersections each have size 19, the third two-way intersection has size 9, and the three-way intersection has size 5. This makes the sum 105 - 39 - 39 - 49 + 19 + 19 + 9 - 5 = 20.

• Question 3 (25): This problem deals with two recurrences. The first has rule b_n = 2b_n-1 + 3b_n-2, with initial conditions b₀ = 1 and b₁ = -1. The second has rule a_n = 2a_n-1 + 3a_n-2 + 16n - 36, with initial conditions a₀ = -3 and a₁ = 3.
  • (a, 5) Find the values of a₁ and b_i for all i with 0 ≤ i ≤ 5.

    a₂ = -7, a₃ = 7, a₄ = 21, a₅ = 107. b₂ = b₄ = 1, b₃ = b₅ = -1.

  • (b, 5) Find a general solution for recurrences with the rule of {b_n}, and find the specific solution with the given initial conditions.

    The characteristic polynomial is α² - 2α - 3 = 0, which has roots 3 and -1. So the general solution is A3ⁿ + B(-1)ⁿ. Substituting b₀ gives A + B = 1, and b₁ gives 3A - B = -1. Adding these two equations gives 4A = 0, so A = 0, B = 1, and the specific solution is b_n = (-1)ⁿ.

  • (c, 5) Find a particular solution for the inhomogeneous part of the recurrence {a_n}.

    We know that the specific solution will be of the form a_n = B₁n + B₀. We have that B₁n + B₀ = 2(B₁(n-1) + B₀) + 3(B₁(n-2) + B₀) + 16n - 36. Equating the terms in n gives B₁ = 5B₁ + 16, which solves to B₁ = -4. Equating the constant terms gives B₀ = -8B₁ + 5B₀ - 36 which gives -4B₀ = -4 or B₀ = 1. So the specific solution is a_n = -4n + 1.

  • (d, 10) Find a general solution for recurrences with the rule of {a_n}, and find the specifc solution with the given initial conditions.

    Combining the answers above, the general solution is a_n = A3ⁿ + B(-1)ⁿ - 4n + 1. Substituting a₀ gives -3 = A + B - 0 + 1, or A + B = -4. Substituting a₁ gives 3 = 3A - B - 4 + 1, or 3A - B = 6. Adding the equations gives 4A = 2 or A = 1/2, and thus B = -9/2.

• Question 4 (15): For any integer n with n ≥ 0, let f(n) = ∑ⁿ_i=0∑ⁱ_j=0 (-2)^i-jC(n, i)C(i, j). Prove that for all positive n, f(n) = 0. (On the exam as given, it asked to prove this for all non-negative n, but it happens to be false for n = 0.)

  We first pull the C(n, i) term out of the inner sum since it is invariant under j. We now have, for each i, the binomial expansion of (1-2)ⁱ. The sum for all i of C(n, i)(-1)ⁱ is just the binomial expansion of (1-1)ⁿ, which is 1 for n = 0 and 0 for all positive integer n.

  It's natural to try to prove this result by induction. The base case of n=1 works fine: We have three terms in the sum, for (i, j) = (0, 0), (1, 0), and (1, 1), and these are (-2)⁰C(1, 0)C(0, 0), (-2)¹C(1, 1)C(1, 0), and (-2)⁰C(1, 1)C(1, 1), giving 1-2+1 = 0.

  So we assume that the given sum is 0, and look at the sum for i from 0 to n+1 of the sum for j from 0 to i of (-2)^i-jC(n, i)C(i, j). Most of you who attempted this proof made a key mistake at this point. None of the terms of the sum in the IH occur in the sum in the inductive goal. The former all have a term of C(n, i), while the latter have C(n+1, i). We indeed add n+2 new terms, for the case of i = n+1. This gives us the sum for j from 0 to n+1 of (-2)^n+1-jC(n+1, n+1)C(n+1, i), which we might recognize as the binomial expansion of (1-2)ⁿ⁺¹ = (-1)ⁿ⁺¹.

  But we also get, for each (i, j) with 0 ≤ i ≤ n and 0 ≤ j ≤ i, a term for the difference between the (i, j) terms in the two sums. By Pascal's Identity, C(n+1, i) - C(n, i) is equal to C(n, i-1), so we have the sum for all such i and j of (-2)^i-jC(n, i-1)C(i, j). The only way I can see to evaluate this is to follow the method I used above to solve the original problem! We pull the C(n, i-1) out of the inner sum, making it the sum for j from 0 to i of (-2)^i-jC(i, j), and this is (1-2)ⁱ. Thus the outer sum becomes the sum for i from 0 to n of C(n, i-1)(-1)ⁱ. We can rewrite this by letting k = i-i to get the sum for k from 0 to n-1 of C(n, k)(-1)^k+1. (We ignore the zero term for k = -1.) If this sum were from 0 to n, we would have (-1) times the binomial expansion of (1-1)ⁿ, which is zero. So our sum is actually -1 times the missing k = n term, and that missing term is C(n, n)(-1)ⁿ⁺¹.

  So the sum for the changed terms exactly cancels the sum for the new terms, and the sum in the inductive goal has the same value as the sum in the IH, which we assume to be 0.

  Again, I didn't expect you to carry out this inductive proof, and no one did so, but I went through it to show that if you didn't make the mistake I pointed out, it was at least theoretically possible.

Last modified 2 December 2016