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smmc29
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GMAT 1: 700 Q49 V35 (Online)
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01 Mar 2021, 13:24
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This will be easier to solve if we assume that at least 1 couple sits together and then subtract the probability of at least 1 couple sitting together from 1.
At least 1 couple:
Consider that couple as 1 individual. Therefore we have 4 remaining sits to be filled = 4!
But the couple may sit in 2 ways
Total number of ways of at least 1 couple sitting together = 2*4! = 48
Both couples sitting together:
Consider couple 1 as 1 individual, and couple 2 as 1 individual. There are then 3! ways to arrange them.
Now per couple we have 2! ways of arranging them. Therefore 3!2!2! = 24
Total ways where at least 1 couple sits with each other = 48 + 24 = 72.
Total number of ways the 5 individuals can be arranged = 5! = 120.
Probability that neither couples sit with each other = 1 - 72/120 = 2/5.
At least 1 couple:
Consider that couple as 1 individual. Therefore we have 4 remaining sits to be filled = 4!
But the couple may sit in 2 ways
Total number of ways of at least 1 couple sitting together = 2*4! = 48
Both couples sitting together:
Consider couple 1 as 1 individual, and couple 2 as 1 individual. There are then 3! ways to arrange them.
Now per couple we have 2! ways of arranging them. Therefore 3!2!2! = 24
Total ways where at least 1 couple sits with each other = 48 + 24 = 72.
Total number of ways the 5 individuals can be arranged = 5! = 120.
Probability that neither couples sit with each other = 1 - 72/120 = 2/5.
19 May 2023, 04:01
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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
19 May 2023, 22:42
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Elite097
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.
