Question

In how many ways 15 chocolates can be distributed among 3 boys so that each one gets at least one chocolate no two boys gets equal number of chocolates.

• A. 30
• B. 60
• C. 72
• D. 30

Answer 1

The correct option is B

60

Let the boys get a, b and c chocolates

$a,b,c\ge 1$ {a, b and c are distinct}

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

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

Number of solution is equal to coefficient of $x^{15}$ in

$(x^3+x^6+x^9+...........)(x^2+x^4+x^6+............)(x+x^2+x^3+.................)$

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

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

Since three distinct 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 = $12\times 6=72$