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Combination/Probability

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Combination/Probability

by MBALA2009 » Thu Jul 10, 2008 9:49 pm
There are 8 students. 4 of them are men and 4 of them are women. If 4 students are selected from the 8 students, what is the probability that the number of men is equal to the number of women?

A) 18/35
B) 16/35
C) 14/35
D) 13/35
E) 12/35
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by lightbulb » Thu Jul 10, 2008 10:00 pm
4M, 4W

Need: 2M, 2W

Favorable ways: C(4,2) * C(4,2) = 6*6 = 36
Total ways: C(8,4) = 70

Pr = 36/70 = 18/35

Ans: A
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by sudhir3127 » Fri Jul 11, 2008 4:01 am
IMO answer is A,

probability of choosing 2 boys out of 4 is 4C2 'lly girls 4C2.

probability of chossing 4 students out of 8 is 8C4.

hence its.. (4C2 x 4C2)/8C4 = 36/70= 18/35.

Sudhir
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How can you do this in 2 minutes?

by evansbd » Fri Jul 11, 2008 11:28 am
Can anyone suggest a way to do 8C4 by hand quickly?

4C2 is simple enough if you remember factorials well....but the others, not so much.


1! - 1
2! - 2
3! - 6
4! - 24
5! - 120
6! - 720
7! - 5040
8! - 40320
9! - 362880
10! - 3628800
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by BlueRain » Sun Jul 13, 2008 10:42 pm
8C4 = 8! / [4! x (8-4)!]
=8! / 4! / 4!
=8x7x6x5x4x3x2x1 / 4x3x2x1 / 4! (cancel 4x3x2x1 terms in 8!)
=8x7x6x5 / 4x3x2x1 (cancel 4x2 with 8, divide 6 by 3 leaves 2)
=7x2x5
=70

Most of the terms in permutation/combinations cancel due to sharing same tail terms (1x2x3xetc.), so you don't have to do the whole 8!.

Hopefully this helps.
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