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permutations & combinations

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by mkhanna » Sun Aug 30, 2009 8:16 am
Q) A committee of 3 people is to be chosen from 4 married couples. What is the number of different committees that can be chosen if 2 people who are married to each other cannot serve on the committee?

A) 16
B) 24
C) 26
D) 30
E) 32
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by tom4lax » Sun Aug 30, 2009 8:24 am
IMO E

8*6*4 = 192 / 3! = 32
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by mkhanna » Sun Aug 30, 2009 8:36 am
That's right. Can you help me understand the logic behind your answer?


Thanks!
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by tom4lax » Sun Aug 30, 2009 8:49 am
Sure. We have 8 ppl, grouped as follows:

A1 A2 B1 B2 C1 C2 D1 D2

These 8 people are to be placed into 3 "slots"

Slot 1 = A1 (or any of the above 8) so slot 1 is 8 choices

Slot 2 = Must not be A2, but can be any of the others, so slot 2 is 6 choices. But for purposes here we say we pick B1

Slot 3 = Must not be A2 or B2, so we have 4 remaning choices, so slot 3 is 4.

8*6*4 = 192. ABC can be arranged in 3! ways (order here does not matter). So divide by 24.

192/24=32
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by mkhanna » Mon Aug 31, 2009 8:29 am
Alright. Understood. Thanks!
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by mkhanna » Mon Aug 31, 2009 8:30 am
Hey..any suggestions for prepping for permutations and combinations and probability..its a particularly weak area for me....thanks!!
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