Three boys and two girls stand in a queue. The probability, that the number of boys ahead of every girl is at least one more than the number of girls ahead of her, is
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Sol. 3 Boys & 2 Girls..................
B B B
Girl can't occupy 4 th position. Either girls can occupy 2 of 1, 2, 3 position or they can both be a position or .
Hence total number of ways in which girls can be seated is 3 C 2 × 2! × 3! + 2 C 1 × 2! × 3! = 36 + 24 = 60.
Number of ways in which 3 B & 2 A can be seated = 5 !
Hence required prob. = 
= 
.
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