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 $\frac{1}{25!}$)?

kindly advise. Regards

  • 1
    $\begingroup$ Two things happened in that step. You noticed the numerators of all the little fractions. But what happened to the numerator of the big fraction? $\endgroup$
    – Arthur
    Apr 7, 2022 at 7:46
  • 3
    $\begingroup$ It's just dividing the numerator and denominatir by $26!$ $\endgroup$ Apr 7, 2022 at 7:52
  • 1
    $\begingroup$ Well, what is the one special thing you learn from early on that you can do to fractions, which doesn't change their value but changes what they look like? Does it look like maybe that's what has happened here? $\endgroup$
    – Arthur
    Apr 7, 2022 at 7:52
  • 2
    $\begingroup$ $\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}} $ $\endgroup$
    – Henry
    Apr 7, 2022 at 8:02
  • 1
    $\begingroup$ It is just distribution and cancelation of common factors. $$\dfrac{26!}{(26!\,a+\cdots+26!\,z)}=\require{cancel}\dfrac{\cancel{26!}}{\cancel{26!}}\dfrac{1}{(a+\cdots+z)}$$ $\endgroup$ Apr 7, 2022 at 8:02


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