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Combinatorics

by Shridharvk » Mon Nov 01, 2010 8:54 am
Q) A park has 4 chairs a,b,c and d. 3 women and 3 men come to the park. In how many ways can 4 people be seated such that the first 2 chairs are occupied by women?

Please note that this is a problem created by me! So I don't know what answer options to give.
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by kmittal82 » Mon Nov 01, 2010 9:05 am
a and b must seat women only.

so, we have to chose 2 women out of 3 to sit on those. This can be done in 3C2 = 3 ways

This leaves 4 people (3 men + 1 woman) to be seated on c and d, which can be done in 4C2 = 6 ways

Thus, total ways = 3 x 6 = 18.

OA?
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by DarkKnight » Mon Nov 01, 2010 11:30 am
I am getting 72 (3*2*4*3).

Whats the OA?
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by rros0770 » Mon Nov 01, 2010 12:56 pm
I came up with 72 also. I interpretted it as a permutation question because it appears Shridharvk wants the seating order to matter based on how it was written out.

If order doesn't matter the Kmittal's correct with 18.


Shridharvk, being that you made up this question, which way did you mean for it to be interpretted?
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by Rezinka » Mon Nov 01, 2010 9:06 pm
I am getting 72 as well.
Here's my logic :

2 women can be selected froom 3 for the first 2 chairs in 3C2 ways.
These 2 women can be arranged on 2 chairs in 2! ways.
For the rest of the two chairs, we have 4 people (1W, 3M) and we can choose any 2 in 4C2 ways.
They can be arranged among themselves in 2! ways.

So, total no. of ways = 3C2 * 2! * 4C2 * 2!
= 3 * 2 * 6 * 2
= 72
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