Question

In how many ways can $14$ identical toys be distributed among $3$ boys so that each one gets atleast one toy and no two boys get equal number of toys ?

• A. $10$
• B. $20$
• C. $60$
• D. none of these

Answer 1

The correct option is C $60$

Let the boys get $a,b$ and $c$ toys.
$\therefore a+b+c=14(\because a,b,c\ge 1,$
and $a,b,c$ are distinct$.)$
$\text{Let}a<b<c\text{and}{x}_1=a,x_2=b-a,x_3=c-b$
$\therefore 3x_1+2x_2+x_3=14,x_1,x_2,x_3\ge 1$
$\text{Therefore the number of solution is equal to coeff. of}{t}^{14}\text{in}(t^3+t^6+t^9+\cdots )\times (t^2+t^4+\cdots )(t+t^2+\cdots )$
$=\text{coeff. of}{t}^8\text{in}(1+t^3+t^6)(1+t^2+t^4+t^6+t^8)\times (1+t+t^2+...+t^8)$
(neglecting higher powers )
$=\text{coeff. of}{t}^8\text{in}$
$(1+t^2+t^3+t^4+t^5+2t^6+t^7+2t^8)\times (1+t+t^2+....+t^8)$
$=1+1+1+1+1+2+1+2$
$=10$
Now three distinct number can be assigned to three boys in $3!$ $(=6)$ ways.
so corrosponding to each solution,we have $6$ ways of distribution.
$\text{So total number of ways}=10\times 6=60$