Dimension of an antisymmetric tensor product space

Question

can somebody explain to me why the dimension of an antisymmetric tensor product space $\Lambda^{r} V$ of rank $r$ and formed from a vector space $V$ with, $\quad dim V = n \quad$ is $\quad {n \choose r}$ ?

Answer 1

Choose a basis of $V$, say $\{e_1,\ldots,e_n\}$. A basis of $\Lambda^r V$ is $\{e_{i_1}\wedge \cdots \wedge e_{i_r}\}_{1\leqslant i_1 < \ldots < i_r \leqslant n}$. Thus the dimension is the number of $r$-tuples in a set of $n$ elements : ${n}\choose{r}$

Answer 2

Take a basis $e_1, \dots, e_n$ of $V$. A non-zero $k$-vector is of the form $$e_{i_1} \wedge \dots \wedge e_{i_k}.$$ Since $\wedge$ is skew-symmetric, the order of the factors does not matter, and you have $n$-elements to choose from to construct your $k$-vector. So you have a combination of $k$ elements out of $n$; which is $\binom{n}{k}$.