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Two couples and one single person are seated at random in a row of fiv [#permalink]
KarishmaB pls restate the double counting here in this approach mentioned by you and please tell me how to account for that and proceed with this approach. Im clear that the base we have is 3!*2*2/5! and now we need to multiply this by somethign
KarishmaB wrote:
Elite097 wrote:
avigutman KarishmaB i approached it differently like i placed one of each couple and the single person already using 2c1 2c1 3!. Then for the other patner i saw how many spaces they have available Ex _ c1_ c2 _ s _ so each of the partners had 2 seats to choose from and hence 2c1 2c1 thus we have 2 2 3! 2 2 / 5!= 4/5. Im not sure where am i wrong . Pls suggest


H1, W1, H2, W2, S

Say in 2C1 * 2C1, you selected H1, W2 and arranged them in 3! ways one of them being H1, W2, S.
Then you said that both W1 and H2 have two possible spots. But one of those spots is common to both (after S). What happens if W1 is placed after S? Then there is one option left for H2.

So how many options we have for a person would depend on where the other people are placed before them.
Also there will be double counting.

That is why questions involving "cannot sit together" are done by making them sit together.

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GMAT Club Bot
Two couples and one single person are seated at random in a row of fiv [#permalink]
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