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A committee of 6 is chosen from 8 men and 5 women so as to

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A committee of 6 is chosen from 8 men and 5 women so as to [#permalink] New post 10 Sep 2004, 01:43
A committee of 6 is chosen from 8 men and 5 women so as to contain at least 2 men and3 women . How many different committees could be formed if two of the men refuse to serve together ?

(A) 3510
(B) 2620
(C) 1404
(D) 700
(E) 635
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 [#permalink] New post 10 Sep 2004, 06:31
This one has beeen already solved here :

either 2 M and 4W or 3M and 3W

so 5C3(8C3-6)+5C4(8C2-1) = 635 if I remember well
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Re: Combination - Committee [#permalink] New post 10 Sep 2004, 08:01
I think it is D.

It can be 3M,3W or 2M,4W

=> (8c3 * 5C3) + (8C2*5C4) = 700

I am not understanding why E is the answer.

Correct me if I am wrong.

rahul wrote:
A committee of 6 is chosen from 8 men and 5 women so as to contain at least 2 men and3 women . How many different committees could be formed if two of the men refuse to serve together ?

(A) 3510
(B) 2620
(C) 1404
(D) 700
(E) 635

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Awaiting response,

Thnx & Rgds,
Chandra

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 [#permalink] New post 10 Sep 2004, 21:48
can some one post details explaination .

OA is E
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 [#permalink] New post 11 Sep 2004, 00:51
Let me try..
1. It can be (2M, 4W) or (3M,3W)
2. The restraint is 2 of the men refuse to serve togethe. So, if we select 2M - then the 2M can be a)1 of the 2Men who dont get together and the other man can be one of the members from the remaining 6 men.

Also, if we choose 3M there can be a) 1 of the two Men who dont get along and the remaining two Men from the other remaining 6 Men

Also, in both the above situations it can be that b)we dont choose either of the 2M and choose the 2M or 3M from only the remaining 6 M.

3. Having said the above - lets put the combinations

Scenario I: 2M, 4W => a)2C1*6C1*5C4 + b)6C2*5C4 =>135
Scenario II: 3M, 3W =>a)2C1*6C2*5C3 + b)6C3*5C3 =>500

Total 135 + 500 = 635. Ans. E.
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 [#permalink] New post 11 Sep 2004, 01:12
This is a detailed explanation of how to approach such a problem step by step.

Let's start from scratch.
The only ways you can form the mentioned committee are with 3men+3women or 2men+4women
# of ways to select 3 women: 5C3 = 10
# of ways to select 3 men out of 8: 8C3 = 56
6C1 comes from this graph:
Y-A-(X-X-X-X-X-X)
Here is the unfavorable outcome whereby A does not want to sit with Y. Let's say A sits with Y, an unfavorable outcome, there are then 6C1 ways of selecting the third person from the remaining 6 men in the group.
Total # of ways of selecting 3 men out of way with the restriction that 2 do not want to sit together is: 8C3 - 6C1 = 56-6 = 50
10*50=500

The same reasoning goes for when you have a committee of 4 women and 2 men
# of ways of selecting 4 women out of 5: 5C4 = 5
# of ways of selecting 2 men out of 8, 2 of which do not want to sit together: Y-A-(X-X-X-X-X-X) --> A does not want to sit with Y, in how many ways can you select 0 persons from the group of 6 men remaining? 1 way. You only want to select 2 men anyways. Therefore, you do not need a third one from the group of remaining 6 men. :btw nC0 = 1. There is always 1 way of selecting nothing from a group of n objects. Therefore, 8C2 - 6C0 = 28 - 1 = 27
5*27 = 135

In the end, you add up all favorable outcomes which are 500+135=635
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 [#permalink] New post 11 Sep 2004, 23:00
First solve the question without any restriction. then apply the restriction.
Next use the POE.

For this question
Without restriction
8C3 x 5C3 + 8C4 x 5C2 = 700

so straight away eliminate A B C and D leaving E as the only choice
  [#permalink] 11 Sep 2004, 23:00
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