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

Options

A. 220
B. 180
C. 140
D. 80

Answer

C

Explanation

Let `x_(1),x_(2),x_(3)andx_(4)` be the numbers on the upper faces of the four dice. Then the required number of ways is equal to the dice. Then the required numbre of ways is equal to the number of solutions of
`x_(1)+x_(2)+x_(3)+x_(4)=13, "where"1lex_(1),x_(2),x_(3),x_(4)le6`
The number of solutions of this equation is equal to the
Coefficient of `x^(13)" in "(x^(1)+x^(2)+....+x^(6))^(6)`
= Coefficient of `x^(13)" in "x^(4)(x+1+x+x^(2)+....+x^(5))^(4)`
= Coefficient of `x^(9)" in "((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)+....)(1-x)^(-4)`
= Coefficient of `x^(9)" in """^(4)C_(0)(1-x)^(-4)`
- Coefficient of `x^(9)" in """^(4)C_(1)x^(6)(1-x)^(-4)`
`""^(4)C_(0)xx("Coefficicent of"x^(9)" in "(1-x)^(-4)`
`-""^(4)C_(0)xx("Coefficicent of "x^(3)" in "(1-x)^(-4)`
`=""^(4)C_(0)xx""^(9+4-1)C_(4-1)-""^(4)C_(1)xx""^(3+4-1)C_(4-1)`
`=""^(12)C_(3)-4xx""^(6)C_(3)=220-80=140`.