Question

In how many different ways ,can $3$ persons $A,B,C$ having $6$ one rupee coin, $7$ one rupee coin, $8$ one rupee coin, respectively donate $10$ one rupee coin collectively ?

• A. $66$
• B. $56$
• C. $47$
• D. $50$

Answer 1

The correct option is C $47$

The number of ways in which they can donate $10$ rupee is same as the number of solution to the equation
$x_1+x_2+x_3=10$
$\text{subjected to the condition :}$
$0\le x_1\le 6,0\le x_2\le 7,0\le x_3\le 8;$
therefore required number of ways
$=\text{coeff. of}{x}^{10}\text{in}(1+x+x^2+\cdots +x^6)(1+x+x^2+\cdots +x^7)(1+x+x^2+\cdots +x^8)$
$=\text{coeff. of}{x}^{10}\text{in}\frac{(1-x^7)}{(1-x)}\cdot \frac{(1-x^8)}{(1-x)}\cdot \frac{(1-x^9)}{(1-x)}$
$=\text{coeffi. of}{x}^{10}\text{in}(1-x^7)(1-x^8)(1-x^9)(1-x{)}^{-3}$
$=\text{coeff. of}{x}^{10}\text{in}(1-x^7-x^8-x^9+x^{15})\times (1+{}^3{C}_{1}x+{}^4{C}_2{x}^2+{}^5{C}_2{x}^3+\cdots +{}^{12}{C}_2{x}^{10}+\cdots )$
$={}^{12}{C}_2-{}^5{C}_3-{}^4{C}_2-{}^3{C}_1$
$=66-10-6-3$
$=47$