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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 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
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 [#permalink] New post 08 Jan 2005, 19:39
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
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 [#permalink] New post 09 Jan 2005, 09:37
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..
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 [#permalink] New post 09 Jan 2005, 10:07
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
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 [#permalink] New post 09 Jan 2005, 10:25
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.
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 [#permalink] New post 23 Jan 2005, 19:05
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
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 [#permalink] New post 23 Jan 2005, 23:33
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.
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 [#permalink] New post 24 Jan 2005, 09:31
excellent method banerjeea

thanks
  [#permalink] 24 Jan 2005, 09:31
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