# $\frac{26!}{\sum_{k=1}^{26} \binom{26}{k}k!}$ trouble understanding how factorials are factored and how it's derives to Taylor series

I am trying to understand how a probability function is derived to Taylor series and how the factorials were factored to the final answer.

$$\frac{26!}{\sum_{k=1}^{26} \binom{26}{k}k!}\\ = \frac{26!}{ \sum_{k=1}^{26} \frac{26!}{k!(26-k)!}k!}\\ = \frac{26!}{\frac{26!}{25!}+\frac{26!}{24!}+\cdots +\frac{26!}{1!}}$$

I can understand the content of the above functions, however, i can't understand how it derives to:

$$= \frac{1}{\frac{1}{25!}+\frac{1}{24!}+\cdots+\frac{1}{1!}+1}$$

Why are all the numerators (26!) of the denominators changed to 1 (examples: $$\frac{26!}{25!}$$ changed to $$\frac{1}{25!}$$)?

• It's just dividing the numerator and denominatir by $26!$ Apr 7, 2022 at 7:52
• $\dfrac{a}{\frac{a}{b}+\frac{a}{c}+\frac{a}{d}} = \dfrac{a}{a\left(\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right)} = \dfrac{1}{\frac{1}{b}+\frac{1}{c}+\frac{1}{d}}$ Apr 7, 2022 at 8:02
• It is just distribution and cancelation of common factors. $$\dfrac{26!}{(26!\,a+\cdots+26!\,z)}=\require{cancel}\dfrac{\cancel{26!}}{\cancel{26!}}\dfrac{1}{(a+\cdots+z)}$$ Apr 7, 2022 at 8:02