Combinatorics: Lecture Notes

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MATH3143 Combinatorics. Lecture Notes (2012), Week 1

Chapter I: Permutations, combinations, occupancy problems. We start with some basic counting principles and examples. I.1. Two ways of counting the same finite set give the same answer. Example I.2. (Hand shaking lemma). The number of delegates at a conference who shake hands an odd number of times is even. Proof. Let D1 , ..., Dn be the delegates, and let X = {(i, j) : Di and Dj shake hands}, and let k = |X|. We count k in two ways. First k is even because it is twice the number of handshakes. Secondly k = k1 + ... + kn where ki = the number of times that Di shakes hands. THUS k1 + .. + kn is even, which implies that evenly many of the ki are odd, which is what we wanted to prove. I.3. (Multiplication principle.) If we make k successive choices, and for 1 ≤ i ≤ k the number of ways of making the ith choice given the previous choices is ni . THEN total number of choices is n1 × ... × nk . I.4. For k ≤ n the number of ways of arranging (or permuting, or selecting in order) k out of n objects, is n × (n − 1) × .. × (n − k + 1) = n!/(n − k)!. This number is also denoted P (n, k). Example I.5. In a race with 20 horses the number of ways in which the first 3 places can be filled is 20 × 19 × 18 = P (20, 3) = 6840. I.6. Note that P (n, n) = n! = the number of permutations of a set of n objects. I.7. The number of ways of choosing or selecting k out of n objects (without regard to order) is denoted C(n, k) (C stands for “combination”). This is 1

the same as the number of k element subsets of an n-element set. Moreover C(n, k) = P (n, k)/k! = n!/(n − k)!k!. Proof. Each choice of k out of n objects gives rise to k! permutations of k out of n objects. So C(n, k).k! = P (n, k). ETC. Note C(n, k) = C(n, n − k). (Why? Because to choose k out of n is the same thing as choosing the complement.) Example I.8. The number of poker hands drawn from a normal pack of cards is C(52, 5) = P (52, 5)/5!. I.9. The number of subsets...
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