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likewhoa: Senior | Next Rank: 100 Posts; Posts: 49; Joined: Sun Aug 12, 2007 8:17 pm

by likewhoa » Fri Nov 23, 2007 12:32 am

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

Answer: D

Can anyone explain?

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Source: Beat The GMAT — Problem Solving | Fri Nov 23, 2007 12:32 am

sujaysolanki: Master | Next Rank: 500 Posts; Posts: 214; Joined: Wed Nov 14, 2007 6:30 am; Thanked: 15 times

by sujaysolanki » Fri Nov 23, 2007 1:45 am

5 couples say a b c d e

2 2 2 2 2

Now if we apply the condition then we can select one from each in 5C3 ways ..so i think it shd be 5c3 * 2 = 20

Please correct me if i am wrong

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samsonite: Newbie | Next Rank: 10 Posts; Posts: 2; Joined: Sun Oct 21, 2007 2:26 am

Solution

by samsonite » Fri Nov 23, 2007 4:55 am

1. Selecting the first member - There are 10 ways of selecting the first person out of a total of 10 people.
2. Selecting the second member - Now you are left with only 8 members to select from because the spouse of the already selected member is not eligible.
3. Selecting the last member - Again the spouse of the 2nd member is not eligible so you are left with only 6 people to choose from.

So the total number of ways the committee can be formed is:
10*8*6 = 480.

However, the three members within each group can be arranged in 3! ways (meaning there are 6 occurences of the same group in the total).

Hence the total number of distinct groups that you can select is 480/6 = 80.

Regards,
Samsonite

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samsonite: Newbie | Next Rank: 10 Posts; Posts: 2; Joined: Sun Oct 21, 2007 2:26 am

Another way of solving this question

by samsonite » Fri Nov 23, 2007 5:27 am

Number of ways of picking the first group = 5
Number of ways of picking a member from this group = 2

Number of ways of picking the second group = 4
Number of ways of picking a member from this group = 2

Number of ways of picking the last group = 3
Number of ways of picking a member from this group = 2

Total number of arrangements = 5*2*4*2*3*2 = 480
Number of redundant groups (because H1W2H3 is the same as W2H1H3) = 3! = 6.

Hence the number of ways of picking 3 members is 480/6 = 80.

Remember: Combination = Permutation/redundancy

Regards,
Samsonite

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sujaysolanki: Master | Next Rank: 500 Posts; Posts: 214; Joined: Wed Nov 14, 2007 6:30 am; Thanked: 15 times

by sujaysolanki » Fri Nov 23, 2007 8:21 am

Neat explanation ...point noted ..

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Orwell_Jetski: Newbie | Next Rank: 10 Posts; Posts: 5; Joined: Sat Nov 03, 2007 12:53 pm

by Orwell_Jetski » Fri Nov 23, 2007 12:46 pm

Thanks guys!

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