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The number of ways in which 4 boys and 3 girls can be seated in a row such that girls and boys are alternate is ::

Answer given by ME is $4!*3! = 144$,  but i think $\binom 5 3*4!*3!$ Should be correct.

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First Boys are arranged in 4! ways

now we have 3 positions for girls So, 3C3

B_B_B_B

 Now girls can be swap their seats therefore 3C3* 3!

so total , 3!*4!
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