Evaluating $\sum_{k=a}^{15} \frac{k!}{(k-a)!}$ for $0 \leq a \leq 15$ I am trying to simplify the sum $\sum_{k=a}^{15} \frac{k!}{(k-a)!}$ for a given integer $a \in [0, 15]$.
Plugged into WolframAlpha, I see that the expression is equivalent to $\frac{16!}{(a+1) \cdot (15 - a)!}$ but I have no clue as to how to get there.
Any help would be appreciated!
 A: We're essentially adding up the numbers along a diagonal of Pascal's triangle since the $a$-th entry of the $k$-th row is $\binom{k}{a}=\frac{k!}{a!(k-a)!}$. And sums of diagonals on Pascal's triangles have some patterns that will give us the expression in question. (Also, it's common to use $n$ for the row number and $k$ for the horizontal positition; so marking the row with $k$ might be a bit confusing for some people).
Anyways, lets define
$$D_k(a,b)=\sum_{n=a}^{b}\binom{n}{k}$$
This is just the sum of the $a$-th through the $b$-th number along the $k$-th diagonal of Pascal's triangle. (For clarity, I'm using the traditional variables; so the $a$ and $k$ here are different than the ones in your question -- I'll clear it up before the end though!)
Because each entry of Pascal's triangle is the sum of the two above it, $\binom{n}{k}=\binom{n-1}{k}+\binom{n-1}{k-1}$, the function $D_k(a,b)$ also satisfies a similar equality:
$$D_k(a,b)=D_{k}(a-1,b-1)+D_{k-1}(a-1,b-1)$$
We can rearrange and use some nice cancellation to get a nice expression for $D_k(a,b)$:
$$D_{k-1}(a-1,b-1)=D_k(a,b)-D_k(a-1,b-1)=\binom{b}{k}-\binom{a-1}{k}$$
Adding 1 to the variables and we get
$$D_k(a,b)=\binom{b+1}{k+1}-\binom{a}{k+1}$$
Switching back to your question now, let's call $S_a=\sum_{k=a}^{15}\frac{k!}{(k-a)!}$. With the right fiddling, we get
$$\frac{1}{a!}S_a=\sum_{k=1}^{15}\frac{k!}{a!(k-a)!}=\sum_{k=a}^{15}\binom{k}{a}=D_{a}(a,15)=\binom{16}{a+1}-\binom{a}{a+1}=\frac{16!}{(a+1)!(15-a)!}$$
If that last equality seems like a jump, just notice that $\binom{a}{a+1}$ is actually outside Pascal's triangle and is just zero. But with all that done, we just multiply through by $a!$ to arrive at WolframAlpha's formula:
$$S_a=\frac{a!16!}{(a+1)!(15-a)!}=\frac{16!}{(a+1)(15-a)!}$$
A: In general: $$\sum_{k=a}^n\binom{k}{a}=\binom{n+1}{a+1}$$
(For a proof see here)
As @Jean Marie remarked this is the so-called hockey stick identity.
Applying that we find:$$\sum_{k=a}^{15}\frac{k!}{\left(k-a\right)!}=a!\sum_{k=a}^{15}\binom{k}{a}=a!\binom{16}{a+1}=\frac{16!}{\left(a+1\right)\left(15-a\right)!}$$
