Permutations problem

D

Darya

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Jan 17, 2020
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154
This sounds like an easy problem but I for some reason can't come to the right solution

6 boys and 6 girls are to form dance couples. Among them, there are boys A, B and girls C, D. How many dance couples can they form if A wants to dance with C, and B wants to dance with D?

The number of all possible variations is 6!. Then with regard to the given condition, the number of couples is 1*1*4*3*2*1. What am I missing?
 
I would set A, B, C, and D aside as they are already paired up. This leaves 4 boys and 4 girls to be paired up. The first girl has her choice of 4 boys, the second girl has her choice of 3 boys, the third girls has her choice of 2 boys, and the last girl is left with the last boy.

And so I would conclude that 4! = 24 couples are possible under the given conditions.
 
MarkFL said:
I would set A, B, C, and D aside as they are already paired up. This leaves 4 boys and 4 girls to be paired up. The first girl has her choice of 4 boys, the second girl has her choice of 3 boys, the third girls has her choice of 2 boys, and the last girl is left with the last boy.

And so I would conclude that 4! = 24 couples are possible under the given conditions.
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It's really weird, it says that the correct solution should be 96
 
Yeah, I don't see how 96 could possibly be correct. :)
 
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