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combination/permutation-PR cat - hard one?

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combination/permutation-PR cat - hard one?

by khurram » Mon Apr 21, 2008 8:53 pm
a five member committee is to be formed from a group of five x and 9 y. if the committee must have at least 2 x and 2 y, in how many diff ways can the committee be chosen.

ans is 1200.

I get 3600.

5!/2!3! * 9/2!7! *10=3600

solution says 1200 as not 2 but 3 ! for one group so 1200 as we will have 3 possible arrangements from the groups we choose.

Thanks very much
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by tomato1 » Mon Apr 21, 2008 11:35 pm
Since atleast two members to be chosen from both the goups, therefore 2 combinations can be possible..............

1. choose 2 from group of 5 and 3 from group of 9

= 5C2 * 9C3 = 840

2. choose 3 from group of 5 and 2 from group of 9

= 5C3 * 9C2 = 360

total= 840+360=1200
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by khurram » Tue Apr 22, 2008 6:24 am
Qucik question.

5!/2!*3! * 9!/2!*7!= 360 choosing two from each group. Not 3 from one and two from the other. Or am i making a calculation mistake.
Thanks
khurram
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by khurram » Tue Apr 22, 2008 6:32 am
I think I get it.

For combination, whether we choose 3 or 2 from 5 it is still 10, so choosing 3 from 9 =10*84=840 and choosing 3 from 5 and 2 from 9 is 10*360= total of 360+840=1200. ANS.

Thanks
Khurram
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by gmat765 » Wed Apr 23, 2008 5:31 pm
thanks.
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