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Unformatted text preview: CSE21 FA11 Homework #3 Solutions (10/15/11) 3.1. Prove by induction that n X k =1 k 2 = n ( n + 1)(2 n + 1) 6 Sketch of proof. Step 1. We need to verify the base case for n = 1, i.e. 1 X k =1 k = 1 = 1 (1 + 1) (2 + 1) 6 . Step 2. Proof by induction. We assume that n k =1 k 2 = n ( n +1)(2 n +1) 6 for some fixed n 1. Now we need to prove that it holds for n + 1. Note that n +1 X k =1 k 2 = n X k =1 k 2 + ( n + 1) 2 = n ( n + 1)(2 n + 1) 6 + ( n + 1) 2 , where we use our assumption in the last equation. Since n ( n + 1)(2 n + 1) 6 + ( n + 1) 2 = ( n + 1)( n + 2)(2 n + 3) 6 , the induction step is complete. 3.2. It is desired to form three committees from a group of 3 men and 3 women. (a) In how many ways can this be done? (b) Suppose that each of the committees must contain at least one man and at least one woman. Now how many different committees are possible? Solution. (a) S (6 , 3) (b) 3!. 3.3. A collection of 5 red jellybeans and 10 blue jellybeans is to be distributed to a group of 3 CSE majors and 2 ECE majors. (a) In how many ways can this be done if there are no restrictions?...
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This note was uploaded on 02/02/2012 for the course CSE 21 taught by Professor Graham during the Fall '07 term at UCSD.
 Fall '07
 Graham

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