18.314 Fall 2007 Problem Set 2. Due Thursday, October 4. 1. For n >= k >= 0, show that C(k,k) + C(k+1,k) + C(k+2,k) + ... + C(n,k) = C(n+1,k+1). (Here C(n,k) is "n choose k".) 2. Find a closed expression for the sum of the entries of the Pascal triangle inside the upper n x n rhombus. For example, for n = 3, you need to sum the entries: 1 1 1 1 2 1 . 3 3 . . . 6 . . 3. Find a closed expression for the number of lattice paths P on the plane with the steps (1,0) "right" and (0,1) "up" such that P starts at the point (0,0), ends at the point (2n,n), and P always stays in the area {x >= y} (below the line x=y). 4. Find the number of compositions of 25 into 5 odd parts. 5. Find a bijection between the following 2 sets: A = {compositions of n with all parts equal to 1 or 2} B = {compositions of n+2 with all parts greater than 1}. 6(*). (optional) Find a bijection between the following 2 sets: A = {compositions of n with all parts equal to 1 or 2} C = {compositions of n+1 with all odd parts} 7. Let p(n,k) be the number of partitions of n with all parts greater then or equal to k. For n, k >= 1, show that p(n+k,k) = p(n,k) + p(n+k, k+1). (optional part) Use this formula, a computer, and your favorite tool (maple, mathematica, C, C++, ...) to calculate p(1000,1). 8. Find a closed expression for the number of set partions of the set {1,2,...,3n} into 3 blocks with n elements. For example, for n=1, there is only one such set partition 1|2|3. (We don't care about ordering of the blocks. Say, 2|1|3 is the same thing as 1|2|3.) 9. Find a bijection between partitions of n with all distinct parts and partitions of n with odd parts. 10. For n and k, let R(n,k) be the number of set partions of [n] with k nonempty blocks such that there is no block that contains two adjacent entries i and i+1. For example, R(5,3) = #{13|24|5, 13|25|4, 14|25|3, 14|35|2, 24|35|1, 15|24|3, 135|2|3} = 7. Express the number R(n,k) in terms of the Stirling numbers of the second kind. 11. Prove the following identity for the signless Stirling numbers of the first kind: 1 c(n,1) + 2 c(n,2) + 3 c(n,3) + ... + n c(n,n) = = n! (1 + 1/2 + 1/3 + 1/4 + ... + 1/n)