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Unformatted text preview: Factoring out n instead we have k n k =k n! (n − k )!k ! (n − 1)! =n (n − 1 − (k − 1))!(k − 1)! n−1 =n k−1 Problem 11 (Fermat’s combinatorial identity) We desire to prove the so called Fermat’s combinatorial identity n k
i=k i−1 k−1 + k k−1 + ···+ n−2 k−1 + n−1 k−1 . = Following the hint, consider the integers 1, 2, · · · , n. Then consider subsets of size k from n elements as a sum over i where we consider i to be the largest entry in all the given subsets k−1 of size k . The smallest i can be is k of which there are subsets where when we k−1 add the element k we get a complete subset of size k . The next subset would have k + 1 k k+1 as the largest element of which there are of these. There are subsets k−1 k−1 n−1 with k + 2 as the largest element etc. Finally, we will have sets with n the largest k−1 n element. Summing all of these subsets up gives . k k−1 k−1 Problem 12 (moments of the binomial coeﬃcients) Part (a): Consider n people from which we want to count the total number of committees n of any size with a chairman. For a committee of size k = 1 we have 1 · = n possible 1 n subsets of two people and two choices choices. For a committee of size k = 2 we have 2 n for the person who is the chair. This gives 2 possible choices. For a committee of size 2 n k = 3 we have 3 , etc. Summing all of these possible choices we ﬁnd that the total 3 number of committees with a chair is
k =1 n k . ...
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- Winter '08