Is it a correct equation for permutations with sets of indistinguishable objects?

Question

C(n, r) = P(n, r)!/r! = n!/r!ㆍ(n-r)!

I'll check if the right hand side of the above equation in Theorem 9.5.2 is correct by expanding the left hand side.

$C(n, n_1)ㆍC(n-n_1, n_2)ㆍC(n-n_1-n_2, n_3)\cdots C(n-n_1-n_2- ...-n_{k-1}, n_k)$
$=\frac {n!}{(n-n_1)!n_1!}\cdot \frac {(n-n_1)!}{(n-n_1-n_2)!n_2!}\cdot \frac {(n-n_1-n_2)!}{(n-n_1-n_2-n_3)!n_3!}\cdots \frac {(n-n_1-n_2-\cdots -n_{k-1})!}{(n-n_1-n_2-\cdots -n_{k-1}-n_k)!n_k!}$
= $= \frac {n!}{n_1!n_2!n_3!\cdots n_k! (n-n_1-n_2-\cdots -n_{k-1} \color{red}{-n_k})!}$
$= \frac {n!}{n_1!n_2!n_3!\cdots n_k! (n-n)!}$ $= \frac {n!}{n_1!n_2!n_3!\cdots n_k!}$

So $(n-n_1-n_2-\cdots -n_{k-1})!$ is not removed from the equation. Is there something wrong in my computation, or is it an author's fault to leaving out the factorial?

[edit] I edited the the pointed part. I now understand it.

Answer 1

The "extra" term isn't $n - n_1 - n_2 - \ldots -n_{k-1}$, it's $n - n_1 - n_2 - \ldots -n_{k}$. $n_1 + n_2 + \ldots + n_k = n$ by definition, so $n - n_1 - n_2 - \ldots - n_{k}$ = $n - (n_1 + n_2 + \ldots + n_k) = n - n = 0$, and $0! = 1$.