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alternate problem...

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alternate problem...

by mehaksal » Sun Aug 26, 2012 10:31 am
Find the numbers of ways in which 4 boys and 4 girls can be seated alternatively :
in a row and there is a boy named John and a girl named Susan amongst the group who cannot be put in adjacent seats
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by adthedaddy » Sun Aug 26, 2012 11:59 pm
Hi, I'll give it a try. Let me know if the answer matches with the OA.

As 4 girls and 4 boys are to be seated alternatively, I assume there are 8 positions. (Generally, GMAT provides the positions data specifically).

Approach:
Ways in which John & Susan are Not Adjacent = Total ways (minus) Ways in which they're adjacent

Part-1: We find total ways

Total ways can be BGBGBGBG and GBGBGBGB
4 boys can sit in 4 places in 4! ways and 4 girls can sit in 4 places in 4! ways.
This can be done in 4!*4! for each arrangement of BGBGBGBG and GBGBGBGB i.e. 2*4!*4!= 1152 .... (1)

Part-2: Ways where John and Susan are together
Consider them as a group. Remaining are 3 boys and 3 girls
3 boys and 3 girls can sit in 3!*3! ways

John(J) and Susan(S) can be arranged in following ways -

JS _ _ _ _ _ _ = 3!*3! ways
_ JS _ _ _ _ _ = 3!*3! ways
_ _ JS _ _ _ _ = 3!*3! ways
_ _ _ JS _ _ _ = 3!*3! ways
_ _ _ _ JS _ _ = 3!*3! ways
_ _ _ _ _ JS _ = 3!*3! ways
_ _ _ _ _ _ JS = 3!*3! ways

Total= 7*3!*3!

Similarly, for Susan-John combination another 7*3!*3!

Combining above, total ways in which John & Susan are together= 2*7*3!*3! = 504 ..... (2)

TOTAL WAYS IN WHICH JOHN & SUSAN ARE NOT ADJACENT = EQN (1) minus EQN (2) = 1152 - 504 = 648
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by Jim@StratusPrep » Mon Aug 27, 2012 7:28 am
Nicely done by adthedaddy. One note is that this is unlikely to be a GMAT problem because it is so involved. However, if you understand it, you should be able to tackle anything that comes your way.
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