How can you simplify this expression? $$\frac{(n+1)!}{k! \cdot (n-k+1)!}$$
I don't have any clue. Can $(n-k+1)!$ be simplified into $(n+1)!(k)!$
I am not too familiar with this symbol: !.
 A: This is the binomial coefficient $\binom{n+1}{k}$.
A: The symbol $!$ is the factorial, it is defined as 
$$n! = n\cdot(n-1)\cdot(n-2)\cdots 2\cdot 1 = \prod_{j=1}^n j.$$
The expression $\frac{n!}{k!\cdot(n-k)!}=\binom nk$ is also known as the binomial coefficient. There is a number of interesting properties of this quantity. Most intuitively, it is the number of ways to pick $k$ elements out of a set that contains $n$ elements.
A: A concrete example might be helpful. Let $n = 4$ and $k = 3$. The expression can be expanded to
$$
\frac{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 }{(3 \cdot 2 \cdot 1)(2 \cdot 1)}.
$$
Notice both the expressions in the denominator are shorter factorials, so you can quickly cancel the "tail" of the factorial in the numerator as
$$
\frac{5 \cdot 4}{2 \cdot 1}
$$
or
$$
\frac{5 \cdot 4 \cdot 3}{3 \cdot 2 \cdot 1}.
$$
(Of course, more cancellation can be done with specific numbers, but I'm trying to mimic the general case where $n$ and $k$ are not specified.)
Back to general $n$ and $k$, we can simplify your original expression as either
$$
\frac{(n+1) \cdot n \cdots (k+1)}{(n+1-k)!}
$$
or
$$
\frac{(n+1) \cdot n \cdots (n+2-k)}{k!}.
$$
In the former, I cancelled the $k!$ "tail". In the latter, I cancelled the $(n+1-k)!$ "tail". Without more information about $n$ and $k$, this is the most cancellation one can hope for.
