Question

In how many ways, can we get a sum greater than $17$ by throwing six distinct dice

• A. $6^6-({}^{17}{C}_6-6{}^{11}{C}_5)$
• B. $6^6-({}^{17}{C}_6+6{}^{11}{C}_5)$
• C. $5^6-({}^{17}{C}_6-6{}^{11}{C}_5)$
• D. $5^6-({}^{17}{C}_6+6{}^{11}{C}_5)$

Answer 1

The correct option is A $6^6-({}^{17}{C}_6-6{}^{11}{C}_5)$

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 :
$1+x_2+x_3+...x_6+x_7=17$
$\text{Here},1\le x_i\le 6\text{where}i=1,2,...6\text{and}{x}_7\ge 0$
$\text{Therefore number of solutions}$
$=\text{coeff. of}{x}^{17}\text{in}(x+x^2+\cdots +x^6{)}^6\times (1+x+x^2+\cdots )$
$=\text{coeff. of}{x}^{11}\text{in}(1-x^6{)}^6(1-x{)}^{-7}$
$=\text{coeff. of}{x}^{11}\text{in}(1-6x^6)(1-x{)}^{-7}$
$={}^{17}{C}_6-6\times {}^{11}{C}_5$
$\text{Total number of cases}=6^6$
$\text{Hence, the number of ways to get a sum greater than}17=6^6-({}^{17}{C}_6-6{\times }^{11}{C}_5)$