Discussion

Explanation:

Since the group has to have atleast three boys, number of ways in which a group of five can be formed = (4C3 × yC2) + (4C4 × yC1) = [4 × (y)(y – 1)/2] + [(1)(y)] = 2y(y – 1) + y   

Since each boy in the group gets a ball, total balls distributed = 3[2y(y – 1)] + 4y = 6y2 – 2y

Since total balls distributed= 368; 6y2 – 2= 368

∴ 3y2 – y – 184 = 0

On solving this, y = 8

Hence, option (d).

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