Question

In how many ways the sum of upper faces of four distinct dice can be six?

• A. $10$
• B. $84$
• C. $56$
• D. None of these

Answer 1

The correct option is A $10$

Let the digit appears on the top face of dice is $x_1,x_2,x_3,x_4$
Here, number of required ways is equal to the number of solution of
$x_1+x_2+x_3+x_4=6,x_1,x_2,x_3,x_4\ge 1$

Let $X_1=x_1-1,X_2=x_2-1,X_3=x_3-1,X_4=x_4-1$
$\Rightarrow X_1+X_2+X_3+X_4=6-(1+1+1+1)=2$
where $X_1,X_2,X_3,X_4\ge 0$

So, the number of solution
$={}^{2+4-1}{C}_{4-1}={}^5{C}_3=10$

Alternate Solution:
For sum of outcomes to be $6$ on the dice
the possible quadruplet of outcomes will
$(1,1,1,4)\&(1,1,2,2)$
So number of possible outcome
$=\frac{4!}{3!}+\frac{4!}{2!\times 2!}=10$