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

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Director
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A group of 6 is chosen from 8 men and 5 women so as to [#permalink]

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15 Jul 2006, 02:37
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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

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Director
Joined: 13 Nov 2003
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15 Jul 2006, 02:56
possible combinations 2m 4W and 3M 3W
For Opt1 4 women can be selected in 5 ways, 2 men out of 8 in 28 ways. Remove 1 because of limitation-27 ways.
For opt 1 we have 135 diff groups
Opt2 3 women can be selected in 10 ways, 3 men in 56. Remove 6 because of limitation-50
for opt2 we have 500 diff groups
Then total is 635

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Manager
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22 Jul 2006, 03:13
Sorry, but how did you arrive at 'Remove 6 because of limitation'

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Intern
Joined: 14 May 2006
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22 Jul 2006, 12:59
Group of 6 with limitation can be formed as
2 M + 4 W or
3 m + 3 W
I forgot total no of men and women. So I am using TM= Total men and TW = Total women

Ans : (TM)C2*(TW)C4 + (TM)C3+(TW)C3

Pravin

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CEO
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Schools: Completed at SAID BUSINESS SCHOOL, OXFORD - Class of 2008

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22 Jul 2006, 13:08
xsports wrote:
Sorry, but how did you arrive at 'Remove 6 because of limitation'

For example if 1 and 2 refuse to work together then cases to be eliminated:

123
124
125
126
127
126
i.e 6 cases.
_________________

SAID BUSINESS SCHOOL, OXFORD - MBA CLASS OF 2008

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Director
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22 Jul 2006, 16:24
mahesh004 wrote:
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

E.

We need atleast 2 men and atleast 3 women. Can do this as follows:

1. 2 men + 4 women
2. 3 men + 3 men

For case 1, possibilities =
8C2 X 5C4 = 140
Subtract possibilities where 2 grumpy men are together (2C2 x 6C0) X 5C4 = 5

Case 1: 135

For case 2, poss=
8C3 X 5C3=560
Subtract poss. where 2 grumps are together (2C2 X 6C1) x 5C3 = 60

Case 2: 500

Total = 500 + 135 = 635

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Re: PS - Permutations   [#permalink] 22 Jul 2006, 16:24
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