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# A Committee of 6 is chosen from 8 men and 5 women, so as to

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Manager
Joined: 25 Aug 2004
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A Committee of 6 is chosen from 8 men and 5 women, so as to [#permalink]  08 Jan 2005, 17:34
A Committee of 6 is chosen from 8 men and 5 women, so as to contain at least 2 men and 3 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
VP
Joined: 18 Nov 2004
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E (635)

scenario 1 : 2 men and 4 women

= (8C2 - 1)x5C4 = 27x5 = 135

scenario 2: 3 men and 3 women

= (8C3 - 1x6C1)x 5C3 = 50x10 = 500

Totol = 500 + 135 = 635 ways
Director
Joined: 21 Sep 2004
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banerjeea_98 wrote:
E (635)

scenario 1 : 2 men and 4 women

= (8C2 - 1)x5C4 = 27x5 = 135

scenario 2: 3 men and 3 women

= (8C3 - 1x6C1)x 5C3 = 50x10 = 500

Totol = 500 + 135 = 635 ways

banerjee, i quite didn't get it.. can you please elaborate..
Manager
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It is actually 635.

I do not understand the logic.

I do agree with the two scenarios.

What does the -1 mean in the following

8C2 - 1

8C3 - 1
VP
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In order to get the combinations where 2 men r not together, easier way to do that is to get all combination - combinations when they are always together. In scenario 1 , there is only 1 way you can select those 2 ppl who don't want to be together, so to get combos where they are not together you do ....(8C2 - 1) ways

In 2nd scenario...you figure out the combos where they are together in a group of 3 men, which is 1 way to select them together and 6C1 ways to select the remaining 1 person in that gp of 3 from remainig 6 men. So u get (8C3 - 1x6C1).

Selecting women in both cases r straight combinations. Hope this makes sense.
Director
Joined: 27 Dec 2004
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banerjeea_98 wrote:
E (635)

scenario 1 : 2 men and 4 women

= (8C2 - 1)x5C4 = 27x5 = 135

scenario 2: 3 men and 3 women

= (8C3 - 1x6C1)x 5C3 = 50x10 = 500

Totol = 500 + 135 = 635 ways

I don't get this part please explain. Thanks

scenario 2: 3 men and 3 women

= (8C3 - 1x6C1)x 5C3 = 50x10 = 500
SVP
Joined: 03 Jan 2005
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Good one! :b:

For three men three women, you need to take out the cases the two men (say A and B) are both selected. Since two of them already are picked (A and B) we just need to choose the third man from the rest of six men, so number of choices is 6C1. This is why it is (8C3-6C1)5C3.
Manager
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excellent method banerjeea

thanks
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