hw8 - EECS 203: Homework 8 Solutions Section 5.3 1. (E) 10...

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

View Full Document Right Arrow Icon

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

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: EECS 203: Homework 8 Solutions Section 5.3 1. (E) 10 P(6,6) = 6! = 720 2. (E) 12 a) C(12,3) = 220 since we only need to select the position of the 1s among the possible 12 positions. b) C(12,3) + C(12,2) + C(12,1) + C(12,0) = 220 + 66 + 12 + 1 = 299 c) C(12,3) + C(12,4) + + C(12,12) = 2 12 [C(12,2) + C(12,1) + C(12,0)] = 4017 d) C(12,6) = 924 since we only need to select the position of six 1s, which automatically leaves us with the remaining positions for 0s. 3. (E) 16 C(10,1) + C(10,3) + C(10,5) + C(10,7) + C(10,9) = 512. This is basically choosing all odd sized subsets, which turns out to be half the total elements, that is 2 10 /2 = 1024/2 = 512. 4. (M) 24 We first position the 10 women, giving us P(10,10) possible ways. This creates 11 possible slots between any two women where a man can stand. Since no two men can stand next to each other, each man must be in one of those 11 slots without repetition, i.e. without two men in the same slot leading to two men next to each other, which gives P(11,6). The final answer is P(10,10)P(11,6) = 10!.11!/5! 5. (E) 26 a) C(13,10) = 286 since we are only choosing 10 out of 13 people....
View Full Document

This note was uploaded on 01/28/2010 for the course EECS 203 taught by Professor Yaoyunshi during the Spring '07 term at University of Michigan.

Page1 / 2

hw8 - EECS 203: Homework 8 Solutions Section 5.3 1. (E) 10...

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

View Full Document Right Arrow Icon
Ask a homework question - tutors are online