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# Classic Comb/Perm/Ps

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VP
Joined: 13 Jun 2004
Posts: 1123
Location: London, UK
Schools: Tuck'08
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Kudos [?]: 31 [0], given: 0

Classic Comb/Perm/Ps [#permalink]  08 Sep 2004, 17:27
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions
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
GMAT Club Legend
Joined: 15 Dec 2003
Posts: 4313
Followers: 26

Kudos [?]: 216 [0], given: 0

D) 700
8c2*5c4 + 8c3*5c3 = 28*5 + 56*10 = 140 + 560 = 700
_________________

Best Regards,

Paul

Director
Joined: 20 Jul 2004
Posts: 593
Followers: 1

Kudos [?]: 47 [0], given: 0

Paul wrote:
D) 700
8c2*5c4 + 8c3*5c3 = 28*5 + 56*10 = 140 + 560 = 700

Paul, did u consider that the two guys wont work together?

I got 700 for all possibilities, by above method. Since these two guys wont work together, the number of ways should be lesser than 700 and 635 is the only choice. E.
Director
Joined: 16 Jun 2004
Posts: 893
Followers: 2

Kudos [?]: 19 [0], given: 0

Yep 635. It would be

6C1*2C1*5C4 + 6C2*5C4 + 6C2*2C1*5C3 + 6C3*5C3 =>60+75+300+200 = 635
GMAT Club Legend
Joined: 07 Jul 2004
Posts: 5078
Location: Singapore
Followers: 22

Kudos [?]: 184 [0], given: 0

Combinations can be 2m,4w
3m,3w
Since the least number of men is 2, and the least number of women is 3

For 2m,4w combinations..

Number of possible outcomes for having 4 women chosen from a group of 5 women = 5C4 = 5
Number of possible outcomes for having 2 men chosen from a group of 8 men = 8C2 = 28
But 1 pair refuse to serve together, we we're left with 28-1 = 27 pairs of men

So total number of possible outcomes for 2m4w combinations = 27*5 = 135

For 3m,3w combinations...
Number of possible outcomes for having 3 women chosen from a group of 5 women = 5C3 = 10
Number of possible outcomes for having 3 men chosen from a group of 8 men = 8C3 = 56
But 1 pair of men are fickled-minded, and so we have to remove 6C1 = 6 combinations, to give 56-6 = 50 pairs of men

Total number of possible outcomes for 3m3w combinations = 50*10 = 500

Total = 500+135 = 635
GMAT Club Legend
Joined: 07 Jul 2004
Posts: 5078
Location: Singapore
Followers: 22

Kudos [?]: 184 [0], given: 0

And as a matter of interest, if you're interested in probability, then
Number of possible outcomes = 13C6 = 1716
Number of favorable outcomes = 635

So P(at least 2 men and 3 women, 2 men refusing to serve on the same team) = 635/1716
GMAT Club Legend
Joined: 15 Dec 2003
Posts: 4313
Followers: 26

Kudos [?]: 216 [0], given: 0

hardworker_indian wrote:
Paul wrote:
D) 700
8c2*5c4 + 8c3*5c3 = 28*5 + 56*10 = 140 + 560 = 700

Paul, did u consider that the two guys wont work together?

I got 700 for all possibilities, by above method. Since these two guys wont work together, the number of ways should be lesser than 700 and 635 is the only choice. E.

This is pretty bad... I think I skipped the last sentence...
_________________

Best Regards,

Paul

Director
Joined: 20 Jul 2004
Posts: 593
Followers: 1

Kudos [?]: 47 [0], given: 0

Paul wrote:
hardworker_indian wrote:
Paul wrote:
D) 700
8c2*5c4 + 8c3*5c3 = 28*5 + 56*10 = 140 + 560 = 700

Paul, did u consider that the two guys wont work together?

I got 700 for all possibilities, by above method. Since these two guys wont work together, the number of ways should be lesser than 700 and 635 is the only choice. E.

This is pretty bad... I think I skipped the last sentence...

Oh oh, stop staring man. You seem to do this everyday evening in the beach.
I bet someone passed by when you skipped the last senetence.
GMAT Club Legend
Joined: 15 Dec 2003
Posts: 4313
Followers: 26

Kudos [?]: 216 [0], given: 0

I think this is going to mark me in my GMAT quest. intr3pid what have you done to me
_________________

Best Regards,

Paul

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