GMAT Prep Q-24

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sky828i: Junior | Next Rank: 30 Posts; Posts: 11; Joined: Wed Jun 20, 2007 12:00 pm; Location: st.louis

by sky828i » Thu Oct 04, 2007 9:20 am

A committee of three 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?
1. 16
2. 24
3. 26
4. 30
5. 32

can anyone explain this ?

the answer is 32.

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Source: Beat The GMAT — Problem Solving | Thu Oct 04, 2007 9:20 am

mschling52: Master | Next Rank: 500 Posts; Posts: 105; Joined: Mon Sep 18, 2006 12:34 pm; Location: OH; Thanked: 7 times; GMAT Score:780

by mschling52 » Thu Oct 04, 2007 9:37 am

I went about this one by first finding the number of different groups possible without any restrictions and then finding the number that involved a married couple and subtracting.

Number of possible groups

8 people, choose 3 = (8!)/(3!5!) = 56 (from formula for combinations)

Number of possible groups with a married couple

For a given couple, they can be involved together in 6 different groups. This is because they must be 2 of the members and there are then 6 possibilities for the other group member. Or, in terms of combinations

Groups with married couple = (2 C 2)(6 C 1) = 1*6 = 6

This is true of all 4 married couples, so in total there are 4*6 = 24 group combinations that involve a married couple. Subtracting 24 from 56 gives us 32 groups that will not involve a couple.

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raulverde: Senior | Next Rank: 100 Posts; Posts: 91; Joined: Wed May 16, 2007 5:46 am; Thanked: 2 times

by raulverde » Thu Oct 04, 2007 8:54 pm

Lets say the couple are as follows :

Male 1 , Female 1
Male 2 , Female 2
Male 3 , Female 3
Male 4 , Female 4

For the 1st Member of the committee we have 8 choices (Assume we picked Male 1)

For the 2nd Member of the committee we have 6 choices (Already picked Male 1 and cant pick Female 1. Assume this time we picked Female 3 )

For the 3rd Member of the committee we have 4 choices (Already picked Male 1 and Female 3. Cant pick Female 1 and Male 3)

Now the committee can be arranged in 3! ways = 6

So the different committee are : (8*6*4)/6 = 32

HTH

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