Math3143 Week 1 Question Paper
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. …show more content…
× 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 of an n element set is 2n . Proof. Let the set be {a1 , .., an }. A subset S of this set is determined by choosing or each i whether ai ∈ S or ai ∈ S. By the …show more content…
Proof. Let {a1 , .., an } be our set of n objects. A selection of k from these, without regard for order, and with repetitions allowed, is the SAME thing as the choice of nonnegative integers x1 , .., xn where x1 is the number of times a1 is chosen, x2 is the number if times a2 is chosen etc., and note that x1 +...+xn has to equal k. Lemma I.15. The number of solutions to x1 + ... + xn = k where xi are nonnegative integers is C(n + k − 1, n − 1). Proof. Fix n + k − 1 spaces in order. Suppose we choose n − 1 of these spaces on which to put markers. Let x1 be the number of spaces before (but not including) the first marker, x2 the number of spaces between the first and second marker, ..., xn the number of spaces after the (n − 1)th marker. Then x1 , .., xn are nonnegative integers whose sum is k (Why??) Conversely, given such x1 , .., xn gives in the same way a choice of n − 1 spaces out of the n + k − 1 spaces for the markers. Hence the number of such (x1 , .., xn ) is the number of ways of choosing n − 1 out of n + k − 1 which equals C(n + k − 1, n − 1) = C(n + k − 1, k). Lemmas I.14 and I.15 yield Theorem I.13. So we see that the number of ways of placing 5 indistinguishable balls in 3 boxes, is the number of ways of choosing 5 out of 3, without regard to order and with repetitions allowed is C(3 + 5 − 1, 2) = C(7, 2) = 7!/(5!.2!) = (7.6)/2 = 21. (So to apply Theorem I.14 we take n = 3 and k = 5.) A final interesting remark, connected with the first part of this
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. …show more content…
× 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 of an n element set is 2n . Proof. Let the set be {a1 , .., an }. A subset S of this set is determined by choosing or each i whether ai ∈ S or ai ∈ S. By the …show more content…
Proof. Let {a1 , .., an } be our set of n objects. A selection of k from these, without regard for order, and with repetitions allowed, is the SAME thing as the choice of nonnegative integers x1 , .., xn where x1 is the number of times a1 is chosen, x2 is the number if times a2 is chosen etc., and note that x1 +...+xn has to equal k. Lemma I.15. The number of solutions to x1 + ... + xn = k where xi are nonnegative integers is C(n + k − 1, n − 1). Proof. Fix n + k − 1 spaces in order. Suppose we choose n − 1 of these spaces on which to put markers. Let x1 be the number of spaces before (but not including) the first marker, x2 the number of spaces between the first and second marker, ..., xn the number of spaces after the (n − 1)th marker. Then x1 , .., xn are nonnegative integers whose sum is k (Why??) Conversely, given such x1 , .., xn gives in the same way a choice of n − 1 spaces out of the n + k − 1 spaces for the markers. Hence the number of such (x1 , .., xn ) is the number of ways of choosing n − 1 out of n + k − 1 which equals C(n + k − 1, n − 1) = C(n + k − 1, k). Lemmas I.14 and I.15 yield Theorem I.13. So we see that the number of ways of placing 5 indistinguishable balls in 3 boxes, is the number of ways of choosing 5 out of 3, without regard to order and with repetitions allowed is C(3 + 5 − 1, 2) = C(7, 2) = 7!/(5!.2!) = (7.6)/2 = 21. (So to apply Theorem I.14 we take n = 3 and k = 5.) A final interesting remark, connected with the first part of this