Question

If $4$ dice are rolled, then the number of ways of getting the sum $10$ is

• A. $56$
• B. $64$
• C. $72$
• D. $80$

Answer 1

Answer: (D)

$80$

As atleast 1 would be seen while

tossing dices, first giving all four dices

1, sum now becomes 6 which is to be

distributed in 4 dices (Beggar's problem)

No. of ways would be ${}^{6+4-1}{C}_{4-1}{=}^9{C}_3$

$=\frac{9\times 8\times 7}{3\times 2}=84$, but there would be 4

cases when all 6 would fall on one of

die (on that dice 7 would not be possible)

$\Rightarrow$ Total ways = $84-4=80\Rightarrow \left(D\right)$