Mathematics - Permutation & Combination Question with Solution | TestHub

Let $x_1{x}_2{x}_3{x}_4{x}_5=2310=3\cdot 7\cdot 10\cdot 11$. Each of 3, 7, 10, 11 can be distributed at 5 places in 5 ways.

Positive integral solutions are $5^5$.

i) If two are negative and three are positive, there are ${}^5{\mathrm{C}}_2\cdot 5^5$ ways.

ii) If four are negative and one is positive, there are ${}^5{\mathrm{C}}_4\cdot 5^5$ ways.

Therefore, the total number of ways is $5^5{(}^5{\mathrm{C}}_0{+}^5{\mathrm{C}}_2{+}^5{\mathrm{C}}_4)=5^5(1+10+5)=16\cdot 5^5$.