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Counting Number of k-tuples

Let $V$ be a vector space of dimension $n$ and let $\Lambda^{k} (V)$ denote the set of all $k$-forms on $V$. If $\{v_1, ..., v_n\}$ is a basis for V, and I denotes the set of all increasing multi-indices of length $k$, it follows that

$$ \{v_{i_1} \wedge, \dots, \wedge v_{i_k} \; | \; (i_1, \dots, i_k) \in I \} $$

is a basis for $\Lambda^{k} (V)$. The dimension of $\Lambda^{k} (V)$ is of course the number of elements in a basis and this is well-known to be given by $dim(\Lambda^{k} (V)) =$ $n \choose{k}$.

I am trying to understand how to provide a rigorous proof of the fact, from first principles, that $dim(\Lambda^{k} (V)) =$ $n \choose{k}$. The combinatorial aspects however are giving me difficulties.Can someone point me in the right direction? Please, assume that I know absolutely nothing about combinatorics (which is an understatement).

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    I hope it is straightforward that these vectors span $\Lambda^k(V)$; this just uses linearity and antisymmetry. The nontrivial step is to show that they are linearly independent and this question is addressed in the linked question.2011-10-04
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    @QiaochuYuan I understand how to show that the indicated set is a basis. What I don't understand is how to actually *count* them and demonstrate that their number is $n \choose k$2011-10-04
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    Sorry, I misunderstood your question. (To be honest, though, I don't think it was stated clearly.) Okay, so increasing multi-indices of length $k$ are in bijection with subsets of $\{ 1, 2, ... n \}$ of size $k$, of which there are ${n \choose k}$. Is it the first statement you're having trouble with or the second?2011-10-04
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    @Qiaochu: this question asks «why are those $\binom n k$ elements of $\Lambda^kV$ linearly independent?». There is nothing in the linked question about that!2011-10-04
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    A TeX thing: use `{n \choose k}` (with braces!) or `\binom{n}{k}`2011-10-04
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    @Mariano I imagine that the OP understands why that particular set of elements is a basis, but has trouble counting the number of elements in it. (If not, perhaps the OP should clarify what exactly the question is :).)2011-10-04

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