By David Guichard

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N, then some box labeled i contains at least ri objects. Proof. Suppose not. Then the total number of objects in the boxes is at most (r1 − 1) + n (r2 − 1) + (r3 − 1) + · · · + (rn − 1) = ( i=1 ri ) − n < X, a contradiction. 7 Suppose r > 0 and X ≥ n(r − 1) + 1 objects are placed into n boxes. Then some box contains at least r objects. Proof. Apply the previous theorem with ri = r for all i. • • • Here is a simple application of the Pigeonhole Principle that leads to many interesting questions.

Partition 2[n] as in the first part of the proof. Suppose that A is a subset of the elements of a one or two element chain C, that is, a chain consisting solely of a set S1 of size n/2, if n is even, or of sets S1 and S2 of sizes n/2 and n/2 , with A ⊆ S1 ⊆ S2 , if n is odd. Then no member of C is in the anti-chain. Thus, the largest possible size for an n anti-chain containing A is n/2 − 1. If A is not a subset of the elements of such a short chain, we now prove that there is another chain partition of 2[n] that does have this property.

Find the number of permutations of 1, 2, . . , 8 that have at least one odd number in the correct position. 6. How many permutations of [n] have exactly k numbers in their correct positions? 7. Give a combinatorial proof that n n! = k=0 n Dn−k . k 8. A small merry-go-round has 8 seats occupied by 8 children. In how many ways can the children change places so that no child sits behind the same child as on the first ride? The seats do not matter, only the relative positions of the children. 9. On the way into a party everyone checks a coat and a bag at the door.

### An Introduction to Combinatorics and Graph Theory by David Guichard

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