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A committee of 3 people is to be chosen from four married

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Re: permutation problem [#permalink] New post 06 Nov 2012, 12:23
breakit wrote:
Bunuel wrote:
breakit wrote:
4C2*2=6*2=12. ??? (4*3*2*1)/(2*1) = 12 but you have mentioned as 6.. is something I am missing here.


C^2_4=\frac{4!}{2!2!}=6.




C^2_4=\frac{4!}{2!2!}=6.


sorry , my mind is totally screwed with above stuff.. help me here[/quote]

Please check here: math-combinatorics-87345.html

Hope it helps.
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Re: A committee of 3 people is to be chosen from four married [#permalink] New post 06 Nov 2012, 12:53
is this a correct way to get the answer? or was it just coincidence:

8C3 - 4(4C2) = 56 - 4(6) = 32
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Re: A committee of 3 people is to be chosen from four married [#permalink] New post 07 Nov 2012, 05:37
watwazdaquestion wrote:
is this a correct way to get the answer? or was it just coincidence:

8C3 - 4(4C2) = 56 - 4(6) = 32


It's not clear what is the logic behind the formula.

Reversed approach would be:
There are 8C3=56 ways to select 3 people out of 8 without any restriction;
There are 4C1*6=24 ways there to be a couple among 3 members: 4C1 ways to select a couple out of 4, which will be in the committee and 6 ways to select the third remaining member (since there will be 6 members left after we select a couple out of 8 people).

56-24=32.

Hope it's clear.
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Re: A committee of 3 people is to be chosen from four married [#permalink] New post 27 Dec 2012, 21:54
LM wrote:
A committee of 3 people is to be chosen from four married couples. What is the number of different committees that can be chosen if two people who are married to each other cannot both serve on the committee?

A. 16
B. 24
C. 26
D. 30
E. 32



How many ways to select 3 represented couples from 4 couples? 4!/3!1! = 4
How many ways to select a person from a pair? 2
=4 * 2 * 2 * 2 = 32

Answer: E

More detailed explanation here : Selection/Deselection Technique
Re: A committee of 3 people is to be chosen from four married   [#permalink] 27 Dec 2012, 21:54
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