Fractional Factorials

$\large = \dfrac{(10!\color{#20A900}{+}9!)(8!\color{#20A900}{+}7!)(6!\color{#20A900}{+}5!)(4!\color{#20A900}{+}3!)(2!\color{#20A900}{+}1!)}{(10!\color{#D61F06}{-}9!)(8!\color{#D61F06}{-}7!)(6!\color{#D61F06}{-}5!)(4!\color{#D61F06}{-}3!)(2!\color{#D61F06}{-}1!)}$

$\large = \dfrac{9!(10 \color{#20A900}{+}1)\cdot 7!(8 \color{#20A900}{+}1)\cdot5!(6 \color{#20A900}{+}1 )\cdot3!(4 \color{#20A900}{+} 1)\cdot1!(2\color{#20A900}{+}1)}{9!(10 \color{#D61F06}{-}1)\cdot7!(8 \color{#D61F06}{-}1)\cdot5!(6 \color{#D61F06}{-}1 )\cdot3!(4 \color{#D61F06}{-}1)\cdot1!(2\color{#D61F06}{-}1)}$

$\large = \dfrac{(10\color{#20A900}{+}1)\cdot (8 \color{#20A900}{+}1)\cdot(6\color{#20A900}{+} 1 )\cdot(4 \color{#20A900}{+}1)\cdot(2\color{#20A900}{+}1)}{(10 \color{#D61F06}{-}1)\cdot(8 \color{#D61F06}{-}1)\cdot(6 \color{#D61F06}{-}1 )\cdot(4 \color{#D61F06}{-}1)\cdot(2\color{#D61F06}{-}1)}$

$\large = \dfrac{11 \times 9 \times 7 \times 5 \times 3}{9 \times \times 7 \times 5 \times 3 \times 1}$

$\large = 11$

It might help to read up on what factorial means. Firstly, note that $10!=10\times 9!$ . This means that $10!+9!=10\times 9! + 9! = 11\times 9!$ and $10!-9!=10\times 9! - 9! = 9\times 9!$ , for example. Therefore, the expression simplifies to:

$\frac{(11\times 9!)(9\times 7!)(7\times 5!)(5\times 3!)(3\times 1!)}{(9\times 9!)(7\times 7!)(5\times 5!)(3\times 3!)(1\times 1!)}$ .

The terms which are present in both the numerator and denominator (e.g. $9!$ , $9$ and $7!$ ) cancel, leaving us with $\frac{11}{1}$ . So the answer is $\boxed{11}$ .

One way to do this faster is to notice that the first term is a multiple of the second term. And then set the second term to a variable and add and subtract regularly and then notice the odd pattern and just cancel out everything except 11.