How can one show $100!=100 \cdot 99!$ by combinatorial arguments How does one show $100!=100\cdot 99!$ by using combinatorial arguments?
 A: $n!$ can represents the numbers of permutations of $n$ different objects in $n$ places. So, if we set $99!$ as the permutation of $99$  different objects in $99$ places, if you add $1$ object and $1$ place then you can choose $99+1=100$ different ways to begin this "sequence" of objects, and then order the $99$ left objects in $99!$ different ways.
More generally, if you have $n$ different objects in $k$ places, you can build $n(n-1)\cdot \dots \cdot(n-k+2)\cdot (n-k+1)$ different ordered sequences, in fact for the first place in the sequence you can choose from $n$ objects, in the second place $n-1$ and so on...
A: Here's one more combinatorial explanation: 
Imagine you have $n$ unique objects (say, diamonds) of which one is particularly important (Arkenstone). You have $n$ slots in a chest to set these diamonds, but you only care about the particular location/slot of Arkenstone. You know there are $n!$ ways of setting these stones, and you absolutely do not care about the order or location of the other $n-1$ stones, obviously there are $(n-1)!$ ways of ordering them. You also know that you can set Arkenstone in any of $n$ slots. Hence,
$$
\frac{n!}{(n-1)!}=n
$$
Division by $(n-1)!$ means you 'do not care' about the order of the other $n-1$ objects. 
A: $$\begin{pmatrix}
100 \\
1 \\
\end{pmatrix} = \frac{100!}{1!(99!)} = 100\Rightarrow 100!=100(99!)$$
