Question

In how many ways can $15$ identical blankets be distributed among $6$ persons such that everyone gets atleast one blanket and two particular persons get equal blankets and another three particular persons get equal blankets.

• A. $8$
• B. $15$
• C. $10$
• D. $12$

Answer 1

The correct option is D $12$

The number of required ways of distributing blankets is equal to the number of solutions of the equation
$3a+2b+c=15(\because a,b,c\ge 1)$
$=\text{coeff. of}{t}^{15}\text{in}(t^3+t^6+t^9+t^{12}\dots )\times (t^2+t^4+...)(t+t^2+...)$
$=\text{coeff. of}{t}^9\text{in}(1+t^3+t^6+t^9)\times (1+t^2+t^4+t^6+t^8)(1+t+t^2+\cdots +t^9)$
( neglecting higher powers )
$=\text{coeff. of}{t}^9\text{in}(1+t^2+t^3+t^4+t^5+2t^6+t^7+2t^8+2t^9)(1+t+t^2+..+t^9)$
$=1+1+1+1+1+2+1+2+2$
$=12$