How to prove that the dimension of exterior p-form is the combinatorial symbol I have read in my book that the dimension of exterior p-form is $$\frac{n!}{p!(n-p)!}$$
Where $n$ is the dimension of the vector space.
But there is no proof of this and I don't know how to prove it.
I know that to solve this problem is equivalent as to determine the number of coefficients $a_{i_1...i_p}$ I have with the constraint that $1<i_k < n$ and $i_1 < ... < i_p$, but I don't know how to compute this.
I would like then a proof that the number of coefficients $a_{i_1...i_p}$ I have under thoose constraints is indeed : $$\frac{n!}{p!(n-p)!}$$
I am not very good in combinatorial proof so I would like a detailed explanation of this if possible.
 A: Basically, a subset of $\{1,\ldots,n\}$ of cardinality $p$ is associated to a unique sequence $1\leqslant i_1<\cdots<i_p\leqslant n$. Therefore, the problem boils down to count the number of subsets of cardinality $p$ of a set of size $n$, which is done in the following manner:


*

*First, you pick an element $k_1$ from $\{1,\ldots,n\}$, there is $n$ choices.

*Then, you pick an element $k_2$ from $\{1,\ldots,n\}\setminus\{k_1\}$, there is $(n-1)$ choices.

*...

*Finally, you pick an element $k_p$ from $\{1,\ldots,n\}\setminus\{k_1,\ldots,k_{p-1}\}$, there is $(n-p+1)$ choices.
You also have to take into account that any permutation of the $k_1,\ldots,k_p$ will give the same set.
Hence, there is $\displaystyle\frac{n(n-1)\cdots(n-p+1)}{p!}$ subsets of $\{1,\ldots,n\}$ having cardinality $p$, which is also: $$\frac{n!}{p!(n-p)!}.$$
A: I think the right constraint is $1 \le i_k \le n$, rather than with "less than" symbols. 
With that in mind, you have $n$ items, and you're choosing exactly $p$ of them. You have $n$ choices for the first, $n-1$ choices for the second, and so on, so there are $n \cdot (n-1) \cdots (n-(p-1))$ ways to pick your items. 
That number is just 
$$
\frac{n \cdot (n-1) \cdots 2 \cdot 1} {(n-p)\cdot(n-p-1)\cdots 2 \cdot 1} = \frac{n!}{(n-p)!}.
$$
Now when you chose those $p$ numbers, you could do it in any order. In choosing two numbers between 1 and 4, you could pick $1$ and then $3$, or $3$ and then $1$, and you'd get the same set of numbers, namely $\{1, 3\}$. 
Because you have $p$ numbers, there are $p!$ orders in which you can shuffle them, so each possible set of numbers has been counted $p!$ times. 
Hence the total number of increasing-index-sets is 
$$
\frac{n!}{(n-p)! p!}.
$$ 
