In how many ways can four persons, each throwing a dice once, make a sum of 13 ?

A
220
B
180
C
140
D
80

Solution:

Let $x_{1}, x_{2}, x_{3}$ be the numbers on the upper faces of the four dice. Then, the required number of ways is equal to the number of solutions of

$x_{1} + x_{2} + x_{3} + x_{4} = 13, where 1 \leq x_{1}, x_{2}, x_{3}, x_{4} \leq 6$

The number of solutions of this equation is equal to the

Coefficient of $x^{13} in {(x^{1} + x^{2} + \dots . + x^{6})}^{4}$

= Coefficient of $x^{9} in {(\frac{1 - x^{6}}{1 - x})}^{4}$

= Coefficient of $x^{9} in {(1 - x^{6})}^{4} (1 - x)^{- 4}$

Coefficient of $x^{9} in (^{4} C_{0} -^{4} C_{1} x^{6} +^{4} C_{2} x^{12} + \dots) (1 - x)^{- 4}$

= Coefficient of $x^{9} {in}^{4} C_{0} (1 - x)^{- 4}$

$=^{4} C_{0} \times^{9 + 4 - 1} C_{4 - 1} -^{4} C_{1} \times^{3 + 4 - 1} C_{4 - 1}$

$=^{12} C_{3} - 4 \times^{6} C_{3} = 220 - 80 = 140$

Solution:

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