Consider the shuffle
$$(f \wedge g)(v_1,\ldots, v_{k + l}) = \sum_{(k,l) - \text{shuffles} \ \sigma} \operatorname{sgn}(\sigma) f(v_{\sigma(1)},\ldots, v_{\sigma (k)})g(v_{\sigma_{(k + 1)}}, \ldots, v_{\sigma_{k + l}})$$
I don't understand why this sum has $k+l\choose k$ terms instead of $(k + l)!$ terms. More specifically how do we know that the count of $\sigma(1) < \sigma(2) < \sigma(3) \ldots < \sigma(k)$ and $\sigma(k + 1) < \ldots < \sigma(k + l)$ has $k+l\choose k$ ways of counting them?