MA 524 Final Fall 2015 Solutions

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1 MA 54 Final Fall 05 Solutions Name: Question Points Score Total: 60

2 MA 54 Solutions Final, Page of 8. Let L be a finite lattice. (a) (5 points) Show that p ( (p r)) (p ) (p r) for all p,, r L. Solution: p p and p (p r) imply p p ( (p r)). p r p and p r (p r) imply p r p ( (p r)). Hence (p ) (p r) p ( (p r)). (b) (5 points) Show that euality holds if L is modular. Solution: If L is modular, then it is graded with rank function ρ. By part (a), it suffices to show that the rank of the two sides is the same. Then ρ(p ( (p r))) = ρ(p) + ρ( (p r)) ρ(p (p r)) = ρ(p) + [ρ() + ρ(p r) ρ( p r)] ρ(p ) = [ρ(p) + ρ() ρ(p )] + ρ(p r) ρ( p r) = ρ(p ) + ρ(p r) ρ((p ) (p r)) = ρ((p ) (p r)). Here we used that and p (p r) = (p (p r)) = p p r = p p r = (p ) (p r).

3 MA 54 Solutions Final, Page of 8. Let P n be the poset with elements x, x,..., x n satisfying x > x < x 3 > x 4 < x 5 > (with all other pairs of elements incomparable). (a) (3 points) Draw the Hasse diagram of J(P 5 ). Solution: (b) (3 points) How many linear extensions does P 5 have? Solution: We wish to count permutations w w w 3 w 4 w 5 S 5 with w > w < w 3 > w 4 < w 5. Clearly either w = or w 4 =. If w =, then there are 4 choices for w ; then w 4 is the minimum remaining number and there are choices for w 3 and w 5 for a total of 8 possibilities. Similarly, there are 8 possibilities if w 4 = for a total of 6 linear extensions. Alternatively, linear extensions of P 5 are in bijection with maximal chains in J(P 5 ). We can find this by labeling the minimum element with and then labeling each element with the sum of the labels of the elements it covers, as shown below

4 MA 54 Solutions Final, Page 3 of 8 (c) (4 points) Show that J(P n ) = f n+, where f n is the nth Fibonacci number (f = f =, f n+ = f n + f n+ for n ). Solution: Since order ideals are in bijection with antichains, we will show that P n has f n+ antichains. Clearly P 0 has f = antichain ( ) while P has f 3 = antichains ( and {x }). For n, any antichain A contains either x n or does not. If it does not, then A is one of the f n+ antichains in P n. If it does, then it cannot contain x n (which is comparable to x n ), so the A\{x n } is one of the f n antichains in P n. Since f n+ = f n + f n+, the result follows. 3. (5 points) Suppose {a n } n satisfies, for all n, Find a d n a d d = n. Solution: By Möbius inversion, a n n = d n µ(d) n/d = d n d n µ(d). Hence a n = d n dµ(d). If n has prime factorization pα pαr r, then µ(d) is only nonzero when d is a product of t distinct p i, in which case µ(d) = ( ) t. Hence a n = p i + p i p j p i p j p k + = ( p i ). i i<j i<j<k i In particular, since 9000 = , a 9000 = ( )( 3)( 5) = 8.

5 MA 54 Solutions Final, Page 4 of 8 4. Let A be the hyperplane arrangement in R 4 consisting of the four hyperplanes x = x, x = x 3, x 3 = x 4, and x 4 = x. (a) (3 points) Draw the Hasse diagram of the intersection poset L(A). Solution: Any pair of hyperplanes intersects in a distinct linear space of dimension : for instance, the first two intersect in the set where x = x = x 3, while the first and third intersect in the set where x = x and x 3 = x 4. However, all four hyperplanes contain the -dimensional subspace x = x = x 3 = x 4. Hence L(A) looks as follows: (b) (4 points) Find the characteristic polynomial χ A (x). Solution: By symmetry, µ(ˆ0, t) for t L(A) only depends on dim t. We find that if dim t = 4, if dim t = 3, µ(ˆ0, t) = if dim t =, 3 if dim t =. (Note t L(A) µ(ˆ0, t) = + 4 ( ) ( 3) = 0.) Hence χ A (x) = µ(ˆ0, t)x dim t = x 4 4x 3 + 6x 3x. t L(A) (c) (3 points) Into how many regions does A divide R 4? Solution: The number of regions of A is ( ) 4 χ A ( ) = = 4.

6 MA 54 Solutions Final, Page 5 of 8 5. (5 points) Let P = 3 4, and let ζ I(P ) be its zeta function. What is ζ 5 (ˆ0, ˆ)? Solution: Recall that ζ 5 (ˆ0, ˆ) counts the number of multichains of length 5 from ˆ0 to ˆ, that is, the number of ways to choose x, x, x 3, x 4 P such that x x x 3 x 4. Since P is a Cartesian product, it suffices to count the number of ways to make these choices in each factor and multiply the answers together. In the chain n, there are (( n 4 )) ( = n+3 ) ways to choose such a multichain. Hence the answer is 4 ( )( )( ) = = (5 points) Ten students are paired up for an assignment. For the next assignment, they must again pair up, but each person must have a different partner than before. In how many ways can this new pairing be made? Solution: Note that the number of ways for k students to pair up is (k )!! = (k )(k 3)(k 5) 3. Indeed, if we order the students, there are k ways for the first student to choose a partner, then k 3 ways for the next unpaired student to choose, and so forth. The number of ways in which a specific m pairs can remain in the new pairing is then (9 m)!!. Since there are ( 5 m) ways to choose these m pairs, by inclusion-exclusion, the number of ways for no pairs to be fixed is ( ) ( ) ( ) ( ) !! 7!! + 5!! 3!! +!! = =

7 MA 54 Solutions Final, Page 6 of 8 7. Let n be a positive integer. (a) (5 points) How many ways are there to write n as a sum of nonnegative integers a, b, c, and d if a must be odd and d must be or 3? Solution: The corresponding ordinary generating function is (x + x 3 + x 5 + )( + x + x + x 3 + ) (x + x 3 x ) = ( x ) ( x) (x + x 3 ) Hence the answer is ( ) n. = x 3 ( + x) ( + x)( x) 3 x 3 = ( x) 3 = ( ) n x n. n 3 (b) (5 points) How many ways are there to write [n] as a disjoint union of sets A, B, C, and D if A must be odd and D must be or 3? Solution: The corresponding exponential generating function is ) ) ( ) (x + ( x3 3! + x5 5! + + x + x! + x3 x 3! +! + x3 3! = ( ) x (ex e x ) e x + x3 6 Hence the coefficient of xn n! is = 4 x (e 3x e x ) + x3 (e 3x e x ) = (3 n x n ) 4 (n )! + (3 n 3 x n ) (n 3)!. n n! 4 (3n ) (n )! + n! (3n 3 ) (n 3)! = n 3 ( ) n 3n + ( ) n 3 3n 3.

8 MA 54 Solutions Final, Page 7 of 8 8. (5 points) Fix positive integers k < n. Let G(k, n) be the set of Grassmannian permutations w S n satisfying w < w < < w k and w k+ < w k+ < < w n. Show that inv(w) = [ ] n. k Solution: For any w G(k, n), note that 0 w < w < < w k (k ) < w k k n k. Hence we may associate to w the partition λ(w) = (w k k, w k (k ),, w, w ) k (n k). This map is a bijection: for any λ = (λ, λ,..., λ k ) k (n k), we define w by w i = λ k+ i + i for i k, and place all other w j (for j > k) in increasing order. Moreover, the number of inversions of w is λ(w) : the only possible inversions occur between w i for i k and w j for j > k, and any w i for i k is contained in exactly w i i inversions (it is inverted with each element of [w i ]\{w i i i}). It follows that [ ] n inv(w) = λ =. k λ k (n k) Solution: Note that there is a bijection between S k S n k G(k, n) and S n : given u S k, v S n k and w G(k, n), we can let u act on the first k letters of w and v act on the last n k letters of w to get a new permutation z S n. To recover (u, v, w) from z, simply sort the first k letters of z (which determines u) and the last n k letters of z (which determines v) to get back w. Note furthermore that inv(z) = inv(u) + inv(v) + inv(w), since any inversion of z either involves two letters among the first k, two letters among the last n k, or one from each group, which correspond to inversions in u, v, or w, respectively. Hence, [n]! = z S n inv(z) = u S k v S n k inv(u)+inv(v)+inv(w) = inv(u) inv(v) u S k v S n k = [k]![n k]! inv(w). Dividing both sides by [k]![n k]! gives the result. inv(w)

9 MA 54 Solutions Final, Page 8 of 8 Solution: Any permutation w G(k, n) has either w k = n or w n = n. Hence If w k = n, then w is obtained from a permutation u G(k, n ) by inserting n into the kth position. Then inv(w) = inv(u) + (n k). If w n = n, then w is obtained from a permutation v G(k, n ) by adding n at the end. Then inv(w) = inv(v). inv(w) = n k u G(k,n ) inv(u) + The result now follows easily by induction using the identity [ ] [ ] [ ] n n n = n k + k k k v G(k,n ). inv(v).

q xk y n k. , provided k < n. (This does not hold for k n.) Give a combinatorial proof of this recurrence by means of a bijective transformation.

q xk y n k. , provided k < n. (This does not hold for k n.) Give a combinatorial proof of this recurrence by means of a bijective transformation. Math 880 Alternative Challenge Problems Fall 2016 A1. Given, n 1, show that: m1 m 2 m = ( ) n+ 1 2 1, where the sum ranges over all positive integer solutions (m 1,..., m ) of m 1 + + m = n. Give both