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

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A group of 6 is chosen from 8 men and 5 women so as to [#permalink] New post 19 Dec 2005, 09:47
A group of 6 is chosen from 8 men and 5 women so as to contain at least 2 men and 3 women. How many different groups can be formed if two of the men refuse to server together?

a) 3510
b) 2620
c) 1404
d) 700
e) 635

I understand the answer listed in Intermediate Math for most part. I don't understand the subtraction part...(how u can account for two of the men refuse to serve together)...can someone elucidate?
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 [#permalink] New post 19 Dec 2005, 13:23
Here is my shot at it

two possibilities exist

i) 2M + 4W and ii)3M and 3W

i) 8x7 x 5x4x3x2 = 6720

ii) 8 x7x6 x 5x4x2 = 20160

2 M do not work together impossible teams

1 x 5x4x3x2 + 1x6 x 5 x4x3 = 120 + 360 = 480

Answer = 6720 + 20160 - 480 = 26400

Not among the choices :(
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 [#permalink] New post 19 Dec 2005, 16:46
I vote for (E) 635.

Total number of cases of making a group of 6 as to contain 2 men and 3 women is 700.

M M M W W W ---> 8C3 * 5C3 = 560
M M W W W W ---> 8C2 * 5C4 = 140

Number of cases of making a group of 6 as to contain 2 men and 3 women, including 2 men (M1 and M2) who don't like each other is 65.

M1 M2 M W W W ---> 6C1 * 5C3 = 60
M1 M2 W W W W ---> 5C4 = 5

--------------------------------------------------------------------------------

700 - 65 = 635.

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 [#permalink] New post 19 Dec 2005, 18:39
Groupings can be (2 men, 4 women) or (3 men, 3 women)


Consider (2men, 4 women)
We have 8 men, but two refuse to work together. Number of ways these 2 men can work together is 1. Total number of ways to pick 2 men from 8 = 8C2 = 28. So number of ways to pick 2 men without two of them in the same team = 27.

Number of ways to pick 4 women = 5C4 = 5.
Total number of combinations for 2men, 4women team = 5 *27 = 135

Consider (3men, 3women)
Number of ways the two men who refuse to work together are picked = 6 (2 of them + match with any 6 others remaining). Number of ways to pick 3 men from = 8C3 = 56. So number of ways to pick 3 men without two of them in the same team = 56-6 = 50

Number of ways to pick 3 women = 5C3 = 10
Total number of combinations for 3men, 3women team = 500

Total number of possibilities = (2men, 4women team) OR (3men, 2 women team) = 135+500 = 635
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 [#permalink] New post 19 Dec 2005, 21:11
gamjatang wrote:
I vote for (E) 635.

Total number of cases of making a group of 6 as to contain 2 men and 3 women is 700.

M M M W W W ---> 8C3 * 5C3 = 560
M M W W W W ---> 8C2 * 5C4 = 140

Number of cases of making a group of 6 as to contain 2 men and 3 women, including 2 men (M1 and M2) who don't like each other is 65.



M1 M2 M W W W ---> 6C1 * 5C3 = 60
M1 M2 W W W W ---> 5C4 = 5

--------------------------------------------------------------------------------

700 - 65 = 635.


agreed E is the answer.

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  [#permalink] 19 Dec 2005, 21:11
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