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I was wondering if there's a formula for the cardinality of the set $A_k=\{(i_1,i_2,\ldots,i_k):1\leq i_1<i_2<\cdots<i_k\leq n\}$ for some $n\in\mathbb{N}$. I calculated that $A_2$ has $n(n-1)/2$ elements, and $A_3=\sum_{j=2}^{n-2}\frac{(n-j)(n-j+1)}{2}$. As you can see, the cardinality of $A_3$ is already represented by a not so nice formula. Is there a general formula? |
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You can also get it by induction using the fairly obvious recurrence $$A_{k+1}(n) = \sum_{i=k}^{n-1}A_k(i):$$ if $A_k(i) = \dbinom{i}{k}$, then $$A_{k+1}(n) = \sum_{i=k}^{n-1}\binom{i}{k} = \binom{n}{k+1}$$ by one of the ‘hockey stick’ identities. |
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The $A_k$ can also be expressed as $\{(i_1,i_2,\ldots,i_k)\;|\; 1\leq i_1\leq n-(k-1),i_1+1\leq i_2\leq n-(k-2),\ldots,i_{k-1}+1\leq i_k\leq n\}$. This way, it is clear how many choices there are for each $i_j$. Multiplying will give you the ol' $n \choose k$ formula. edit: Apologies. It's not clear to me right now how to do the multiplication! |
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