Question

There are total ${}^{17}{C}_6-k{}^{11}{C}_5$ ways to get a sum of atmost $17$ by throwing six distinct dice ,then $k$ is equal to :

• A. $4$
• B. $5$
• C. $6$
• D. $8$

Answer 1

The correct option is C $6$

Let $x_1,x_2\cdots x_6$ be the number that appears on the six dice.
Let us find the number of ways to get the sum less than or equal to $17.$
This will be same as finding the number of solutions to the inequality $x_1+x_2+x_3+...x_6\le 17$
Introducing a dummy variable $x_7(x_7\ge 0),$ the inequality becomes an equation :
$1+x_2+x_3+...x_6+x_7=17$
$\text{Here}1\le x_i\le 6\text{where}i=1,2,...6\text{and}{x}_7\ge 0$
$\text{Therefore the number of solutions}=\text{coeff. of}{x}^{17}\text{in}(x+x^2+...+x^6{)}^6\times (1+x+x^2+...)$
$=\text{coeff. of}{x}^{11}\text{in}(1-x^6{)}^6(1-x{)}^{-7}$
$=\text{coeff. of}{x}^{11}\text{in}(1-6x^6)(1-x{)}^{-7}$
$={}^{17}{C}_6-6\times {}^{11}{C}_5$