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by gmattesttaker2 » Sun Sep 09, 2012 12:16 am
Hello,

Can you please assist with this problem? This is in-class MGMAT problem. Thanks a lot for your help.

Best Regards,
Sri


A committee of 3 people is to be chosen from 4 married couples. What is the number of different married commitees that can be chosen if 2 people who are married to each other can't both serve on the committee.

Answer: [spoiler](8 x 6 x 4)/3! = 32[/spoiler]

My approach was as follows:
H1 W1
H2 W2
H3 W3
H4 W4

So, ( (4 x 3 x 2)/2! ) + ( (4 x 3 x 2)/2! ) = 24
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by GMATGuruNY » Sun Sep 09, 2012 2:47 am
I posted a solution here:

https://www.beatthegmat.com/gmatprep-mar ... tml#305067
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I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
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by Anurag@Gurome » Sun Sep 09, 2012 6:11 pm
gmattesttaker2 wrote:Hello,

Can you please assist with this problem? This is in-class MGMAT problem. Thanks a lot for your help.

Best Regards,
Sri


A committee of 3 people is to be chosen from 4 married couples. What is the number of different married commitees that can be chosen if 2 people who are married to each other can't both serve on the committee.

Answer: [spoiler](8 x 6 x 4)/3! = 32[/spoiler]

My approach was as follows:
H1 W1
H2 W2
H3 W3
H4 W4

So, ( (4 x 3 x 2)/2! ) + ( (4 x 3 x 2)/2! ) = 24
Number of ways to choose 3 people out of 4 = 4C3
Number of different committees that can be chosen if two married people, both cannot serve the committee = 4C3 * 2^3 = 4 * 8 = 32 (2^3, as we can choose any 2 people from each chosen team)

The correct answer is E.
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