HW3_sol

# HW3_sol - CSE21 FA11 Homework #3 Solutions (10/15/11) 3.1....

This preview shows pages 1–2. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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?...
View Full Document

## This note was uploaded on 02/02/2012 for the course CSE 21 taught by Professor Graham during the Fall '07 term at UCSD.

### Page1 / 4

HW3_sol - CSE21 FA11 Homework #3 Solutions (10/15/11) 3.1....

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online