In how many ways can 14 identical toys be distributed among three boys so that each one gets at least one toy and no two boys get equal number of toys.

Explanation

Let the boys get `x_(1),x_(2) " and " x_(3)` toys respectively.
`therefore x_(1)+x_(2)+x_(3)=14, x_(1), x_(2),x_(3) ge 1 " and " x_(1),x_(2) " and " x_(3)` are distinct
let `x_(1) lt x_(2) lt x_(3) " and " x_(2)=x_(1)+alpha, x_(3)=x_(2)+beta, " where " alpha, beta ge 1`
`therefore x_(1)+(x_(1)+alpha)+(x_(1)+alpha+beta)=14`
or `3x_(1)+2alpha+beta=14, x_(1), alpha, beta ge 1`
The number of solutions is equal to the coefficient of `p^(14) " in " (p^(3)+p^(6)+p^(9)..)(p^(2)+p^(4)+..)(p+p^(2)+..)`
= coefficient of `p^(8) " in " (1+p^(3)+p^(6))(1+p^(2)+p^(4)+p^(6)+p^(8))xx(1+p+p^(2)+..+p^(8))`
= coefficient of `p^(8) " in " (1+p^(2)+p^(3)+p^(4)+p^(5)+2p^(6)+p^(7)+2p^(8))xx(1+p+p^(2)+..+p^(8))`
`= 1+1+1+1+1+2+1+2=10`
Since three distinct number can be assigned to three boys in 3! ways.
So, corresponding to each solution, we have six ways of distribution.
Therefore, total numbers of ways `=10xx6=60`.