Permutation/ Combination Question : Quant Question Archive [LOCKED]
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# Permutation/ Combination Question

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10 Dec 2006, 20:37
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Hey guys, I couldn't solve this .........

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 can not both serve on the committee.
1) 16
2) 24
3) 26
4) 30
5) 32
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Maverick

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10 Dec 2006, 20:52
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 can not both serve on the committee.
1) 16
2) 24
3) 26
4) 30
5) 32

Number of ways to form commitee of 3 people 8C3= 8*7*6/3*2= 56
Number of comittes where couples r togather
Assume 1 copl chosen then 1 seat left . number of ways to choose 1 person out of 6
6C1= 6
There are 4 copules so 6*4=24

Answer total - where cpl r togather = 56 - 24= 32
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10 Dec 2006, 22:33
i am getting A...whts the OA?
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11 Dec 2006, 14:17
(4 couples) C 3 = 4
Now each couple has 2 members = 4 * 2 * 2 *2 = 32
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11 Dec 2006, 14:30
8C1 = 8 = No of ways of choosing first person
6C1 = 6 = No of ways of choosing second person (excluding spouse of first)
4C1 = 4 = No of ways of choosing second person (excluding spouse of second)

8*6*4/3*2*1 = 32
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11 Dec 2006, 17:38
8C1 = 8 = No of ways of choosing first person
6C1 = 6 = No of ways of choosing second person (excluding spouse of first)
4C1 = 4 = No of ways of choosing second person (excluding spouse of second)

8*6*4/3*2*1 = 32

Why did you divide it by 3*2*1 ?
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11 Dec 2006, 18:28
trivikram wrote:
8C1 = 8 = No of ways of choosing first person
6C1 = 6 = No of ways of choosing second person (excluding spouse of first)
4C1 = 4 = No of ways of choosing second person (excluding spouse of second)

8*6*4/3*2*1 = 32

Why did you divide it by 3*2*1 ?

8*6*4 = 192 = Total no. of possibilities of a committee of 3 people.

Although we've used the "choose" function to give a unique no. of combinations in each individual case, by multiplying them we have also calculated the way to permute the 3 members.

No of way to permute 3 people = 3*2*1

So dividing by 3*2*1 gives us the number of unique combinations
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11 Dec 2006, 18:47
trivikram wrote:
8C1 = 8 = No of ways of choosing first person
6C1 = 6 = No of ways of choosing second person (excluding spouse of first)
4C1 = 4 = No of ways of choosing second person (excluding spouse of second)

8*6*4/3*2*1 = 32

Why did you divide it by 3*2*1 ?

8*6*4 = 192 = Total no. of possibilities of a committee of 3 people.

Although we've used the "choose" function to give a unique no. of combinations in each individual case, by multiplying them we have also calculated the way to permute the 3 members.

No of way to permute 3 people = 3*2*1

So dividing by 3*2*1 gives us the number of unique combinations

Got it...

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11 Dec 2006, 19:14
Actually, now I'm doubting myself!!

It seems to give the correct answer?!?!?
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11 Dec 2006, 20:42
Thanks a lot guys. The answer is 32. Cheers!
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Maverick

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11 Dec 2006, 21:47
Damager wrote:
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 can not both serve on the committee.
1) 16
2) 24
3) 26
4) 30
5) 32

Number of ways to form commitee of 3 people 8C3= 8*7*6/3*2= 56
Number of comittes where couples r togather
Assume 1 copl chosen then 1 seat left . number of ways to choose 1 person out of 6
6C1= 6
There are 4 copules so 6*4=24

Answer total - where cpl r togather = 56 - 24= 32

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11 Dec 2006, 21:49
trivikram wrote:
8C1 = 8 = No of ways of choosing first person
6C1 = 6 = No of ways of choosing second person (excluding spouse of first)
4C1 = 4 = No of ways of choosing second person (excluding spouse of second)

8*6*4/3*2*1 = 32

Why did you divide it by 3*2*1 ?

8*6*4 = 192 = Total no. of possibilities of a committee of 3 people.

Although we've used the "choose" function to give a unique no. of combinations in each individual case, by multiplying them we have also calculated the way to permute the 3 members.

No of way to permute 3 people = 3*2*1

So dividing by 3*2*1 gives us the number of unique combinations

Didnt get this approach
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14 Dec 2006, 01:56
Another approach:

Total people = 4 * 2 = 8

3 people in committee, so we have 3 positions,

1st position = any of the 8
2nd position = any of the remaining 6 (cant choose the 1st position's spouse)
3rd position = any of the remaining 4

Total possibilities = 8 * 6 * 4

Since order doesn't matter, we need to divide by 3!

So total = (8 * 6 * 4 )/ 6 = 32

belabor wrote:
Hey guys, I couldn't solve this .........

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 can not both serve on the committee.
1) 16
2) 24
3) 26
4) 30
5) 32
Re: Permutation/ Combination Question   [#permalink] 14 Dec 2006, 01:56
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