Question

How many ways we can get a sum of atmost 17 by throwing six distinct dies.

• A. 7756
• B. 9604
• C. 3493
• D. 1914

Answer 1

The correct option is B

9604

Let $x_1,x_2,x_3,x_4,x_5,x_6$ be the number that appears on the six dies.

Here $1\le x_1,x_2,x_3,x_4,x_5,x_6\le 6$

Number of ways to get sum less than or equal to 17

$x_1+x_2+x_3+x_4+x_5+x_6\le 17$

Introducing a dummy variable $x_7(x_7\ge 0)$ inequality becomes an equation

$x_1+x_2+x_3+x_4+x_5+x_6+x_7=17$

$1\le x_1,x_2,x_3,x_4,x_5,x_6\le 6$ and $x_7\ge 0$

Number of solution = Coefficient of $x^{17}$ in $(x+x^2+x^3+.................x^6{)}^6(1+x+x^2+x^3+.............)$

Coefficient of $x^{11}$ in $(x-x^6{)}^6(1-x{)}^{-7}$

Coefficient of $x^{11}$ in $(x-6x^6{)}^6(1-x{)}^{-7}$

Coefficient of $x^{11}$ in $(x-x^6{)}^6(1{-}^7{C}_1{x}^8{C}_2{x}^2{+}^9{C}_3{x}^3+................)$ = ${}^{17}{C}_{11}$ - 6${}^{11}{C}_6$ = ${}^{17}{C}_6$ - 6${}^{11}{C}_5$ = 9604