# sol11(1) - CS 330 Discrete Computational Structures Fall...

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CS 330 : Discrete Computational Structures Fall Semester, 2014 Assignment #11 Solutions [Extra Credit] Due Date: Sunday, Dec 7 Suggested Reading: Rosen Sections 6.4 - 6.5 These are the problems that you need to turn in. Always explain your answers and show your reasoning. Spend time giving a complete solution. You will be graded based on how well you explain your answers. Just correct answers will not be enough! 1. [10 Pts] Peter Rosen, Section 6.4: Exercise 22 Solution: a. Consider a group of n identical figurines to be painted. r of them will be painted purple, and k of those will be painted pink and purple. We can count this in the following ways: Count it by selecting r figurines from n to paint purple, then selecting k figurines from those r to paint pink. This is ( n r )( r k ) . Count it by selecting k figurines from n to paint purple and pink. We now need to paint r - k more figurines purple, so we select r - k from the remaining n - k figurines. This is ( n k )( n - k r - k ) . Therefore, ( n r )( r k ) = ( n k )( n - k r - k ) . b. n r r k = n ! r !( n - r )! × r ! k !( r - k )! = n ! k !( n - r )!( r - k )! = n ! k !( n - r )!( r - k )! × ( n - k )! ( n - k )! = n ! k !( n - k )! × ( n - k )! ( r - k )!( n - r )! = n ! k !( n - k )! × ( n - k )! ( r - k )!(( n - k ) - ( r - k ))! = n k n - k r - k 2. [8 Pts] Peter Prove, using a combinatorial argument, that C ( m + n, 2) = C ( m, 2)+ C ( n, 2)+ mn , where m, n 2. Hint: Consider a group of m men and n women and pick two people from this group. Solution: If we choose 2 people from a group of m + n total people, we can count that as ( m + n 2 ) . If we choose 2 people from a group of m men and n women, either we pick 2 women, or 2 men, or 1 man and 1 woman. There are ( n 2 ) ways to pick 2 women, ( m 2 ) ways to pick 2 men, and mn ways to choose 1 of each, for ( n 2 ) + ( m 2 ) + mn ways to pick 2 people. Therefore, ( m + n 2 ) = ( n 2 ) + ( m 2 ) + mn .

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3. [8 Pts] Shana Rosen, Section 6.4: Exercise 30 Solution: Give a combinatorial proof that n k =1 k ( n k ) 2 = n ( 2 n - 1 n - 1 ) . We show why both the left hand side and right hand side represent the number of ways we can select a committee of n people from a group of n math professors and n computer science professors such that the chairperson of the committee is a math professor. For the right hand side, we select a committee head from the n math professors in n ways. Next, we choose the rest of the committee ( n - 1 people) from the remaining math and cs professors ( n + n - 1 people) in ( 2 n - 1 n - 1 ) ways. This gives n ( 2 n - 1 n - 1 ) . For the left hand side, we can select k members from the math professors in ( n k ) ways and then the remaining n - k members from the computer science professors in ( n n - k ) = ( n k ) ways. Next, we select any of the k math professors chosen to be the group leader in k ways. This gives k ( n k )( n k ) . We can do this process with every value of k from 1 to n , giving n k =1 k ( n k ) 2 ways to pick the committee. Therefore, we have shown that both the left and side and right hand side represent different ways to count the same situation and are equal.
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