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Difficult Math Problem #63 - Combinations

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by 800guy » Fri Nov 24, 2006 11:25 am
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?
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Re: Difficult Math Problem #63 - Combinations

by gmat_enthus » Fri Nov 24, 2006 12:31 pm
800guy wrote: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?
Possibilities:

2m 4w = 8C2*5C4
3m 3w = 8C3*5C3

Total = 8C2*5C4 + 8C3*5C3
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Algebra #62

by lalitaroral » Sat Nov 25, 2006 5:18 am
2m 4w = 8C2*5C4-2C2
3m 3w = 8C3*5C3-3C2

Total = 8C2*5C4 - -2C2 + 8C3*5C3 - 3C2
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by gmat_enthus » Sun Nov 26, 2006 1:31 am
Whats -2C2 for?
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by gmat_enthus » Sun Nov 26, 2006 1:33 am
Whats -2C
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Sorry for typing mistake

by lalitaroral » Sun Nov 26, 2006 3:35 am
possible combination=>

1. 2m 4w
2. 3m 3w

Case 1:

Probabilty of choosing 2m out of 8m = 8C2
Probabilty of choosing 4w out of 5w = 5C4

Total Probability = 8C2*5C4

If the selected two men are the one who doesnt want to be together, Probability of the two men coming together = 2C2 (1 possiblity)

Actual probabilty = 8C2*5C4 - 2C2

Case 2:

Probabilty of choosing 3m out of 8m = 8C3
Probabilty of choosing 3w out of 5w = 5C3

Total Probability = 8C3*5C3

If the two out of three selected men are the one who doesnt want to be together, Probability of the two men coming together = 3C2 (3 possiblity)

Actual probabilty = 8C3*5C3 - 3C2

+++++++++++++++++++++++++++++++++++++++++++
Total possible arrangements => (8C2*5C4 - 2C2)+(8C3*5C3 - 3C2)
+++++++++++++++++++++++++++++++++++++++++++

Please correct me if I am wrong...Looking for a positive feedback...!!!
Just started with my preperations :)
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by gmat_enthus » Sun Nov 26, 2006 8:21 am
Good point, I missed out on "not wanting to be together part".

I think the total number should be

[8C2-2C2]*5C4 + [8C3-3C2]*5C3

Why should we take out 2C2 from the total ways and not from the ways of selecting men?

What do you think?

Thanks
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Perfect

by lalitaroral » Sun Nov 26, 2006 8:47 pm
ahh, I missed that....seems perfect now. ......Thanks for correction!!!
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OA

by 800guy » Mon Nov 27, 2006 12:47 pm
OA:

4W 2M == 5C4.(8C2-1) = 5.(27) = 135
3W 3M == 5C3.(8C3-6) = 10.50 = 500
total 635

Max. number of possibilities considering we can choose any man 8c2 * 5C4 + 8C3*5c3 = 700.
consider it this way.... from my previous reply max possible ways considering we can chose any man = 700

now we know that 2 man could not be together... now think opposite... how many ways are possible to have these two man always chosen together...

since they are always chosen together...

For chosing 2 men and 4 women for the committee there is only 1 way of chosing 2 men for the committee since we know only two specific have to be chosen and there are 5C4 ways of choosing women

1*5C4 = 5

For choosing 3 men and 3 women for the committee there are exactly 6C1 ways choosing 3 men for the committee since we know two specific have to be chosen so from the remaining 6 men we have to chose 1 and there are 5C4 ways of choosing women

6C1*5C3 = 60

So total number of unfavorable cases = 5 + 60 = 65

Now since we want to exclude these 65 cases... final answer is 700-65 = 635
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Re: Difficult Math Problem #63 - Combinations

by real2008 » Sun Jul 12, 2009 11:58 pm
800guy wrote: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?
is the answer 560+140= 700?
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by ladhanivishal » Mon Jul 13, 2009 3:16 am
OA:

4W 2M == 5C4.(8C2-1) = 5.(27) = 135
3W 3M == 5C3.(8C3-6) = 10.50 = 500
total 635
I am getting 665. Why are you subtracting 6 in the second case? It should be 3 according to me.Please explain.Thanks
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Re: OA

by ketkoag » Mon Jul 13, 2009 4:02 am
800guy wrote:OA:

4W 2M == 5C4.(8C2-1) = 5.(27) = 135
3W 3M == 5C3.(8C3-6) = 10.50 = 500
total 635

Max. number of possibilities considering we can choose any man 8c2 * 5C4 + 8C3*5c3 = 700.
consider it this way.... from my previous reply max possible ways considering we can chose any man = 700

now we know that 2 man could not be together... now think opposite... how many ways are possible to have these two man always chosen together...

since they are always chosen together...

For chosing 2 men and 4 women for the committee there is only 1 way of chosing 2 men for the committee since we know only two specific have to be chosen and there are 5C4 ways of choosing women

1*5C4 = 5

For choosing 3 men and 3 women for the committee there are exactly 6C1 ways choosing 3 men for the committee since we know two specific have to be chosen so from the remaining 6 men we have to chose 1 and there are 5C4 ways of choosing women

6C1*5C3 = 60

So total number of unfavorable cases = 5 + 60 = 65

Now since we want to exclude these 65 cases... final answer is 700-65 = 635
Nice question.. I have a query here.. in the bold part above if 2 men are always together then why don't u subtract those 2 from 6 and then select 1 out of remaining 4.. so we have those 2 already + 1 from the remaining 4 men.. so we can select one of them in 4C1 ways.
therefore, we can select 3M + 3W in 4C1 * 5C3 = 40 ways.. instead of 60? right?? please lemme know if i missed something..
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by scoobydooby » Sat Jul 25, 2009 10:11 pm
you only missed there are 8 men in all (not 6 men as per your calculation)

so for the (3M, 3W) combination where both men are selected together, the ways of selection are 1*6C1*5C3=60 ways
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