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# If a committee of 3 people is to be selected from among 5

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If a committee of 3 people is to be selected from among 5 [#permalink]

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01 Dec 2005, 14:33
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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
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01 Dec 2005, 15:03
D. 80

all arrangements:
- first person can be selected from any of 10 people
- second person can be any of remaining 9 except first's partner - 8 ways
- third person similarly can be any of the remaining 8 except 2 partners : 6

all arrangements: 10 * 8 * 6

as we don't need to keep order, selection can be done in:
10*8*6/3! = 80
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01 Dec 2005, 15:24
duttsit wrote:
D. 80

all arrangements:
- first person can be selected from any of 10 people
- second person can be any of remaining 9 except first's partner - 8 ways
- third person similarly can be any of the remaining 8 except 2 partners : 6

all arrangements: 10 * 8 * 6

as we don't need to keep order, selection can be done in:
10*8*6/3! = 80

Duttsit. can you please explain what does "keep order" mean. I cannot understand why we divide by 3!
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01 Dec 2005, 15:30
believe2 wrote:

Duttsit. can you please explain what does "keep order" mean. I cannot understand why we divide by 3!

its a selection problem, we are not concerned about order of selection in committee. So selection of:

A,B,C and
B,A,C

in this order is same(one group of 3 people out of 10)

first part of the explanation found all possible arrangements that include above both.

we need to divide by n! to remove these duplicates.
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01 Dec 2005, 16:40
What level of difficulty would you consider this question to be?
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01 Dec 2005, 17:42
duttsit wrote:
believe2 wrote:

Duttsit. can you please explain what does "keep order" mean. I cannot understand why we divide by 3!

its a selection problem, we are not concerned about order of selection in committee. So selection of:

A,B,C and
B,A,C

in this order is same(one group of 3 people out of 10)

first part of the explanation found all possible arrangements that include above both.

we need to divide by n! to remove these duplicates.

Thanks a lot!
I think my 'combination/selection' logic needs some re-wiring!

I would say it needs two steps to solve so its definately at least a medium level question.
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01 Dec 2005, 19:09
We have 10 people: c1c2, c3c4, c5c6, c7c8, c9c10

For the first slot: we have 10 choices
For the second slot: 8 choices
For the last slot: 6 choices

Total = 10 * 8 * 6 = 480 permutations

We do not need permutation for this case, so total = 480/3!= 80
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01 Dec 2005, 22:09
I did it this way. total number of people is 10, we want to select a committee of 3 . so possibiliteis are 10C3 = 120

now we want to know number of possiblities that don't have people married to each other.

now there are 5 couples, so 5C1*8C1 = 5*8=40 is the possibilites of couples forming the committee.

so number of possiblities that don't have people married to each other = 120-40= 80.

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01 Dec 2005, 22:20
nakib77 wrote:
I did it this way. total number of people is 10, we want to select a committee of 3 . so possibiliteis are 10C3 = 120

now we want to know number of possiblities that don't have people married to each other.

now there are 5 couples, so 5C1*8C1 = 5*8=40 is the possibilites of couples forming the committee.

so number of possiblities that don't have people married to each other = 120-40= 80.

I like this method Nakib. Can you explain where you got the 8C1 from??

Thanx
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01 Dec 2005, 22:34
the desired number of ways of chosing 3 people = total ways - no of ways a couple is selected in the comittee

total = 10C3 = 120
no of ways comittees with a couple is selected = no. of ways 5 couples are selected (no of ways a person is selected from the remaining people
= 5(8C1) = 40

so the desired ways of forming comittee = 120 - 40 = 80
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01 Dec 2005, 23:53
Himalaya,
Can you explain the second part a lil more

How are you getting 5* 8C1
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02 Dec 2005, 00:07
andy_gr8 wrote:
Himalaya, Can you explain the second part a lil more
How are you getting 5* 8C1

take a couple (which you can chose in 5 ways from 5 couples) then chose 1 person from the rest of 8 people in 8C1 or 8 ways. so it is 5 (8C1).

let me know if any?
Re: Can you explain   [#permalink] 02 Dec 2005, 00:07
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