Sol. Let the boys get a, b and c toys respectively

a + b + c = 14, a, b, c ≥ 1 and a, b and c are distinct.

let a < b < c and x 1 = a, x 2 = b – a, x 3 = c – b

So, 3x 1 + 2x 2 + x 3 =14, x 1 , x 2 , x 3 ≥ 1

The number of solutions is equal to the coefficient of t 14 (t

3 + t 6 + t 9 . . .) (t 2 + t 4 +…) (t + t 2 +… )

= Coefficient of t 8 in (1 + t 3 + t 6 ) (1+ t 2 + t 4 + t 6 + t 8 ) (1 +t + t 2 +….+t 8 )

(neglecting higher powers)

= coefficient of t 8 in (1 + t 2 + t 3 + t 4 + t 5 +2t 6 + t 7 + 2t 8 )

(1 + t + t 2 +…..+ t 8 )

= 1 + 1 + 1 + 1 + 1+ 2 + 1 + 2 = 10

Since the three distinct numbers can be assigned to three boys in 3! Ways. So, corresponding to each solution, we have six ways of distribution. So, the total number of ways

= 10 × 6 = 60