Question

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.

Answer 1

Let the boys get $a,b,c$ toys. Now, $a+b+c=14,a,b,c\ge 1$ and $a,b,andc$ are distinct.

Let $a<b<c$ and $x_1=a,x_2=b,x_3=c-b$. So,

$3x_1+2x_2+x_3=13,x_1,x_2,x_3\ge 1$

Therefore, the number of solutions is equal to the coefficient of $t^{14}$ in $(t^3+t^6+t^9+...)(t^2+t^4+...)(t+t^2+...)$

$=$ Coefficient of $t^8$ in $(1+t^3+t^6)(1+t^2+t^4+t^6+t^8)(1+t+t^2+t...+t^8)$ (neglecting higher powers)

$=$ Coefficient of $t^8$ in $(1+t^2+t^3+t^4+t^5+2t^6+t^7+2t^8)\times (1+t+t^2+...+t^8)$

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

Now, three distinct numbers can be assigned to three boys in $3!$ ways.

So, corresponding to each solution, we have six ways of distribution. So, total number of ways is $10\times 6=60$.