Question

In how many ways 15 chocolates can be distributed among 6 children such that everyone gets at least one chocolate and two particular children get equal chocolates and another three particular child gets equal chocolates.

• A. 10
• B. 12
• C. 15
• D. 18

Answer 1

The correct option is B

12

The number of ways of distributing chocolates is equal to number of solution of the equation

$x_1+x_2+x_2+x_3+x_3+x_3=15$

$x_1+2x_2+3x_3=15$ where $x_1,x_2,x_3\ge 1$

Which is equal to Coefficient of $x^{15}$ in $(x+x^2+x^3+..........)(x^2+x^4+x^6+...........)(x^3+x^6+x^9+.............)$

= Coefficient of $x^9$ in $(1+x+x^2+x^3+..............x^9)(1+x^2+x^4+............x^8)(1+x^3+x^6+x^9)$

Neglecting higher powers

Coefficient of $x^9$ in $(1+x+x^2+x^3+.........x^9)(1+x^2+x^3+x^4+x^5+2x^6+x^7+2x^8+2x^9)$

= 1 + 1 + 1 +1 + 1 + 2 + 1 + 2 + 2 = 12

So, number of ways = 12