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This question already has an answer here:

This is a variation of question asked on this site before.

Consider a set with $𝑎_1$ 'distinct' 1s, $𝑎_2$ 'distinct' 2s, ... , $𝑎_𝑛$ 'distinct' ns. You have $𝑎_1+1$ choices for the 1s (including the option of none of them being chosen) and similarly for the other elements. The total number of subsets is therefore $(𝑎_1+1)(𝑎_2+2)...(𝑎_𝑛+1)$

Now how to find the number of subsets with size $\leq $ 'k'.

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marked as duplicate by Mike Earnest, Xander Henderson, nmasanta, John Omielan, José Carlos Santos Sep 8 at 10:36

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We have the multiset $\{ 1^{a_1} ,2^{a_2} , \cdots ,n^{a_n}\}$ (where multiplicity of the elements is indicated by the expononent). The number of subsets (of size $k$) will be the coefficient of $x^k$ in the function below \begin{eqnarray*} (1+x+ \cdots +x^{a_1}) (1+x+ \cdots +x^{a_2}) \cdots (1+x+ \cdots +x^{a_n}). \end{eqnarray*}

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