In Finan's probability book 13.19
A box contains 100 computer mice of which 95 are defective. One mouse is taken from the box at a time (without replacement) until a non-defective mouse is found. Let $X$ be the number of mouses you have to take out in order to find one that is not defective. Find $P(X = n)$.
I thought the way to do this is
${{95!(100-n+1)!}\over{100!(95-n+1)!}}{5\over100-n+1}$
and indeed such is the way it seems from
Repeating something with (1/n)th chance of success n times
see answer from copper.hat there,
but in the book the answer is
$P(X=n)= {95\over100}{94\over99}{93\over98}...{95-n+2\over100-n+2}{5\over100-n+1}.$
Why is it $-n+2$?
When you have two positive integers $k<m$, then $$\frac{m!}{(m-k)!}=\frac{m\cdot(m-1)\cdot(m-2)\cdot\ldots\cdot (m-k+1)\cdot \color{red}{(m-k)\cdot(m-k-1)\cdot(m-k-2)\cdot\ldots\cdot2\cdot 1}}{\color{red}{(m-k)\cdot(m-k-1)\cdot(m-k-2)\cdot\ldots\cdot 2\cdot 1}} =m\cdot(m-1)\cdot(m-2)\cdot\ldots\cdot (m-k+1).$$
Taking $m=95$ and $k=n-1$, we get $${{95!}\over{(95-n+1)!}}=95\cdot 94\cdot 93\cdot\ldots\cdot (95-n+2).$$
Similarly (and taking the multiplicative inverse of both sides) we have $${{(100-n+1)!}\over{100!}}={1\over 100\cdot 99\cdot 98\cdot\ldots\cdot (100-n+2)}.$$
So indeed $$P(X=n)= {95\over100}{94\over99}{93\over98}...{95-n+2\over100-n+2}{5\over100-n+1}={{95!(100-n+1)!}\over{100!(95-n+1)!}}{5\over100-n+1}.$$
Note that the above works for $m>k>0$, but in some way also for $k=0$. If $k=0$, then we simply get $\frac{m!}{m!}=1$. As you can see, in the case $n=1$ (so $k=n-1=0$) the left fractions in the answer of the book simply disappear (just like a sum iterating from $0$ up to $-1$ is equal to $0$, products of nothing are equal to $1$), while your left fraction turns into $1$, so your answer agrees with the one of the book for all $n\geq 1$.