In how many ways can four persons, each throwing a die once, make a sum of 6?A. `""^(9)C_(2)`B. `""^(10)C_(3)`C. `""^(8)C_(3)`D. `""^(9)C_(3)`

In how many ways can four persons, each throwing a die once, make a su

1.	In how many ways can four persons, each throwing a die once, make a sum of 6?A. `""^(9)C_(2)`B. `""^(10)C_(3)`C. `""^(8)C_(3)`D. `""^(9)C_(3)`
Answer» Let, `x_(1),x_(2),x_(3),x_(4)` be the number obtained by the four person on the upper faces of their respective dice. Then, the required number of ways will be equal to the number of solutions of the equation `x_(1)+x_(2)+x_(3)+x_(4)=6,"where"1lex_(1),x_(2),x_(3),x_(4)le6`. Since the upper limit is six which is equal to the sum required. So, the upper limit for each variable can be taken as infinite. Hence, required number of ways = Coefficinet of `x^(6)" in "(1+x+x^(2)+....)^(4)` = Coefficinet of `x^(6)" in "(1-x)^(4)` `=""^(6+4-1)C_(4-1)=""^(9)C_(3)=84`.

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In how many ways can four persons, each throwing a die once, make a sum of 6?A. `""^(9)C_(2)`B. `""^(10)C_(3)`C. `""^(8)C_(3)`D. `""^(9)C_(3)`

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