Trine University Mathematics Competition 2018 Advanced Problem #4

$\begin{aligned} X & = \frac {(10!+9!)(8!+7!)(6!+5!)(4!+3!)(2!+1!)}{(10!-9!)(8!-7!)(6!-5!)(4!-3!)(2!-1!)} & \small \color{#3D99F6} \text{Note that }n! = n(n-1)! \\ & = \frac {(10\cdot 9!+9!)(8\cdot 7!+7!)(6\cdot 5!+5!)(4\cdot 3!+3!)(2\cdot 1!+1!)}{(10\cdot 9!-9!)(8\cdot 7!-7!)(6\cdot 5!-5!)(4\cdot 3!-3!)(2\cdot 1!-1!)} \\ & = \frac {9!(10+1)7!(8+1)5!(6+1)3!(4+1)1!(2+1)}{9!(10-1)7!(8-1)5!(6-1)3!(4-1)1!(2-1)} \\ & = \frac {11\cdot 9 \cdot 7 \cdot 5 \cdot 3}{9 \cdot 7 \cdot 5 \cdot 3 \cdot 1} \\ & = \boxed{11} \end{aligned}$

To begin simplification, know the fact that $10! = 10\times9\times8\times...\times2\times1$ and $9! = 9\times8\times...\times2\times1$ .

Using this knowledge, to simplify $10!+9!$ , we can write $10!$ as $10(9!)$ .

Therefore, $10!+9! = 10(9!)+9! = 11(9!)$ , and $8!+7! = 8(7!)+7! = 9(7!)$ , and so on.

In the denominator, the same concept dictates that $10!-9! = 10(9!)-9! = 9(9!)$ , and $8!-7! = 8(7!)-7! = 7(7!)$ , and so on.

Therefore, the original fraction can be rewritten as $\frac{(11(9!))(9(7!))(7(5!))(5(3!))(3(1!))}{(9(9!))(7(7!))(5(5!))(3(3!))(1(1!))}$ , which can be simplified through cancellation to $11$ .