Problem_Set_05_Solutions

# Problem_Set_05_Solutions - ARSDIGITA UNIVERSITY MONTH 2...

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ARSDIGITA UNIVERSITY MONTH 2: DISCRETE MATHEMATICS PROFESSOR SHAI SIMONSON PROBLEM SET 5 SOLUTIONS — COMBINATORICS AND COUNTING (1) Given ten points in the plane with no three colinear, (a) How many diﬀerent segments joining two points are there? ± 10 2 ² = 45 (b) How many ways are there to choose a directed path of length two through three distinct points? We can choose a directed path of length two uniquely by choosing the starting point, the middle point, and the end point. There are 10 · 9 · 8 = 720 such paths. (c) How many diﬀerent triangles are there? There are ( 10 3 ) = 120 triangles. Note that each triangle corresponds to six directed paths of length two — we can choose which two edges and the direction. (d) How many ways are there to choose 4 segments? ± 45 4 ² = 148 , 995 (e) If you choose 4 segments at random, what is the chance that some three form a triangle? The number of ways to choose four segments that include a triangle is ± 10 3 ² · 42 = 5 , 040 We ﬁrst pick the three vertices that will form the triangle, then choose any one of the 42 remaining segments. The probability that we have picked such ana rrangement when picking four segments at random is therefore 5 , 040 148 , 995 3 . 38%. (2) Forty equally skilled teams play a tournament in which every team plays every other team exactly once, and there are not ties. (a) How many diﬀerent games were played? ± 40 2 ² = 780 (b) How many diﬀerent possible outcomes for these games are there? Each of the 780 games has two possible outcomes, so the total number of diﬀerent outcomes is 2 780 . 1

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2 PROBLEM SET 5 SOLUTIONS (c) How many diﬀerent ways are there for each team to win a diﬀerent number of games? If each team wins a diﬀerent number of games, this corresponds to a unique ordering of the teams. There are 40! such orderings. (3) Let C ( n, k ) be the number of ways to choose k objects from a set of n . Prove by a combinatorial argument: (a) C ( n, 0) + C ( n, 1) + . . . + C ( n, n ) = 2 n On the left hand side, we have the total number of subsets of a set of size n , which is the sum of the number of possible subsets of each size. On the right hand side, we have the size of the power set of a set of n elements. These are equal. (b) C ( n, m ) C ( m, k ) = C ( n, k ) C ( n - k, m - k ) To choose a subset S m of size m and an “inner” subset S k S m of size k from an original set S n of n members, we may proceed by ﬁrst choosing S m , then choosing members of the “inner” subset S k from among S m (the left hand side). Equivalently, we can ﬁrst choose the “inner” subset S k from the original set S n , then forming S m by starting with S k and choosing an additional m - k members from S n \ S k (the right hand side).
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Problem_Set_05_Solutions - ARSDIGITA UNIVERSITY MONTH 2...

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