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A committee of three people is to be chosen from 4 married [#permalink]
18 Nov 2007, 19:15

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This post was BOOKMARKED

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Difficulty:

45% (medium)

Question Stats:

75% (01:54) correct
25% (01:52) wrong based on 55 sessions

A committee of three people is to be chosen from 4 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

Here's my solution so far :

Total combination 8C3 = 56

Choose the first couple 4C1 = 4 Choose the third person= 8C1 = 8 * 7! / 7! = 8

8 * 4 = 32 - Not sure if the answer should be 32 or 24. Please help.

Re: PS-Combinations [#permalink]
17 Jun 2010, 04:50

1

This post received KUDOS

Expert's post

1

This post was BOOKMARKED

amitjash wrote:

Hi everybody, I want to solve this problem with some other method. But i am definately wrong somewhere in this method... i dont understand where.

i can choose first member of the comitee in 8 ways.. removing the spouse of the selected person second member can be chosen in 6 ways.... third member in 4 ways..... so 8*6*4 which is not answer can someone explain why?

The way you are doing is wrong because 8*6*4=192 will contain duplication and to get rid of them you should divide this number by the factorial of the # of people - 3! --> 192/3!=32.

Consider this: there are two couples and we want to choose 2 people not married to each other. Couples: A_1, A_2 and B_1, B_2. Committees possible:

A_1,B_1; A_1,B_2; A_2,B_1; A_2,B_2.

Only 4 such committees are possible.

If we do the way you are doing we'll get: 4*2=8. And to get the right answer we should divide 8 by 2! --> 8/2!=4.

Re: PS - Combination [#permalink]
20 Nov 2007, 12:50

alimad wrote:

A committee of three people is to be chosen from 4 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?

16 24 26 30 32

Here's my solution so far :

Total combination 8C3 = 56

Choose the first couple 4C1 = 4 Choose the third person= 8C1 = 8 * 7! / 7! = 8

8 * 4 = 32 - Not sure if the answer should be 32 or 24. Please help.

following your solution:
Total combination 8C3 = 56
Choose the first couple 4C1 = 4
Choose the third person= 6C1 = 6
so # of comittees with a couple = 4x6 = 24
so # of comittees without a couple = 56 - 24 = 32

A committee of three people is to be chosen from 4 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?

16 24 26 30 32

Soln: Now the first of the 3 people commitee can be chosen from any of the 8 people in 8 ways Now the second of the 3 people commitee can be chosen from any of the 6 people in 6 ways(leaving out the one person who is the couple of the person chosen in the first statement) Now the third of the 3 people commitee can be chosen from any of the 4 people in 4 ways(leaving out the two people whose couples have already been chosen)

Thus the total number of permutations is = 8 * 6 * 4 But since this is a selection, hence order does not matter

Therefore the total number of selections is = 8 * 6 * 4/3! = 32 ways

Sorry...I am not able to move this post to PS section. Could someone please move it to PS section.

Approach I thought is as follows...if some shorter method is possible please explain..

total selections = 8C3 = 56

let's say that couple is always present in this committee of three.

This means that there are 4 ways to select 2 people of the committee. ( 4 couples and any one couple can be selected in 4 ways) The third person can be selected out of remaining 6 people in 6 ways.

Therefore when couple exists there are: 4X6 = 24 ways

Re: PS-Combinations [#permalink]
11 May 2010, 14:24

Expert's post

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

One of the approaches:

Each couple can send only one "representative" to the committee. Let's see in how many ways we can choose 3 couples (as there should be 3 members) out of 4 to send only one "representatives" to the committee: 4C3=4.

But each of these 3 couples can send two persons (husband or wife): 2*2*2=2^3=8.

Re: PS-Combinations [#permalink]
17 Jun 2010, 01:16

Hi everybody, I want to solve this problem with some other method. But i am definately wrong somewhere in this method... i dont understand where.

i can choose first member of the comitee in 8 ways.. removing the spouse of the selected person second member can be chosen in 6 ways.... third member in 4 ways..... so 8*6*4 which is not answer can someone explain why?

The way you are doing is wrong because 8*6*4=192 will contain duplication and to get rid of them you should divide this number by the factorial of the # of people - 3! --> 192/3!=32.

Thanks the two of you!

This is also the way I like to solve such questions and I believe it is way faster than any 10C3... and so on!

Re: A committee of three people is to be chosen from 4 married [#permalink]
12 Aug 2014, 03:54

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