By David Guichard

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Note that by the preceding discussion, 1 ≤ im ≤ k and 1 ≤ jm ≤ k. Then Ajr−s = {x1 , . . , xs , ys+1 , . . , yr−1 , xr }, so Ajr−s ⊇ Bk+1 , a contradiction. Hence there is no such anti-chain. 7. 1. 6) tells us that 63 , with size 20, is the unique largest anti-chain for 2[6] . The next largest anti-chains of the form k6 are 62 and 64 , with size 15. Find a maximal anti-chain with size larger than 15 but less than 20. (As usual, maximal here means that the anti-chain cannot be enlarged simply by adding elements.

There are no solutions with 3 or 4 of the variables larger than 10. Hence the number of solutions is 25 + 3 4 14 + 3 4 3+3 − + = 676. 1. 1. 1, and list the corresponding 6 submultisets. 2. Find the number of integer solutions to x1 + x2 + x3 + x4 = 25, 1 ≤ x1 ≤ 6, 2 ≤ x2 ≤ 8, 0 ≤ x3 ≤ 8, 5 ≤ x4 ≤ 9. 3. Find the number of submultisets of {25 · a, 25 · b, 25 · c, 25 · d} of size 80. 4. 1). Prove that for n k 1 = k! k (−1)k−i in i=0 k . i Do n = 0 as a special case, then use inclusion-exclusion for the rest.

1)k = n! 1 + k=1 n (−1)k = n! k=0 1 k! 1 . k! The last sum should look familiar: ∞ x e = k=0 Substituting x = −1 gives −1 e ∞ = k=0 1 k x . k! 1 (−1)k . k! The probability of getting a derangement by chance is then 1 n! n! n 1 (−1) = k! n k k=0 (−1)k k=0 1 , k! 3678. So in the case of a deck of cards, the probability of a derangement is about 37%. n 1 Let Dn = n! k=0 (−1)k k! These derangement numbers have some interesting properties. The derangements of [n] may be produced as follows: For each i ∈ {2, 3, .