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Question 1 (nov. 25)

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by bacali » Tue Nov 25, 2008 7:50 am
Q3:
If a committee of 3 people is to be selected from among 5 married couples so that the committee does not include two people who are married to each other, how many such committees are possible?


A. 20
B. 40
C. 50
D. 80
E. 120


OA: D
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by mals24 » Tue Nov 25, 2008 8:18 am
You have 5 couples and you want to select 3 people and no 2 people ot of the 3 are married.

We have 10 ways to select the first person.
We have 8 ways to select the 2nd person
We have 6 ways to select the 3rd person.

Since we are just selecting people, ABC is the same as BCA.

Hence to avoid over counting we divide by 3!

Total ways = 10*6*8/3! = 480/3! = 80.
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by bacali » Tue Nov 25, 2008 8:38 am
mals24 wrote:You have 5 couples and you want to select 3 people and no 2 people ot of the 3 are married.

We have 10 ways to select the first person.
We have 8 ways to select the 2nd person
We have 6 ways to select the 3rd person.

Since we are just selecting people, ABC is the same as BCA.

Hence to avoid over counting we divide by 3!

Total ways = 10*6*8/3! = 480/3! = 80.
I was getting the 10*6*8 right but wasn't diving by 3!. Thanks a lot!!
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