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In how many ways can we arrange 4 boys and 3 girls in a straight line such that no two girls are together.

One approach came to my mind was

arrange 4 boys first=4! ways.

Now 5 gaps created.Choose 3 for girls and arrange girls=5C3*3! ways.

Total=4!*5C3*3! ways.

But another approach also came to my mind was

say B1 B2 B3 B4 are boys and G1 G2 and G3 are girls.

total ways to arrange them without any restriction is 7!

now if I want girls to be always together, then I take these 3 girls and single object,these can be permuted with 4 boys in 5! ways and the girls among themselves can be permuted in 3! ways so total 5!*3! ways.

Now if I subtract this quantity from 7!(total ways without restriction) why am I getting a different answer from the former case where I take boys first and then arrange girls?

1 Answer

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You forgot the case where only two girls are together.

Which two girls will be together can be selected in $\binom{3}{2}$ ways. The 5 units (4 boys and 2 girls together) can be permuted in $5!$ ways and the two girls will also permute in $2!$ ways, and for the one separate girl,you will be left with 4 places to select. Hence total number of ways when only two girls are together is:

$\binom{3}{2}*5!*2*4 = 2880$ 

$5040 - 2880 - 720 = 1440  $

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