Q: Six different balls are put in three different boxes, no box being empty. The probability of putting balls in the boxes in equal numbers is,

(A) 3/10

(B) 1/6

(C) 1/5

(D) none of these

Sol. Total number of ways to distribute the balls so that no box is empty are [(1, 1, 4), (2, 1, 3), (2, 2, 2)]

$\large \frac{3!}{2!} [( 6_{C_1} . 5_{C_1} . 4_{C_4} ] + 3! ( 6_{C_2} . 4_{C_1} . 3_{C_3}) + ( 6_{C_2} . 4_{C_2} . 2_{C_2})$

= 90 + 6.60 + 90 = 540

Required probability $\large = \frac{90}{540} = \frac{1}{6} $