# Does fixing the number of elements in PARTITION send it in P?

It's possible that this question is trivial and I've overlooked something.

Let us impose the following constraint on the well-known PARTITION problem: in all inputs of a new problem the number of elements is fixed and equals $k$?

Does such constraint make PARTITION polynomially solvable?

P.S. There is pseudopolynomial algorithm for it with running time $\mathcal{O}(nA)$, where $A$ is the sum of the numbers in the given input. However, fixing $n$ doesn't seem to make it polynomial in $\log A$. Or am I missing something?

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Yes. The problem becomes in $P$. The reason is that there would be a fixed number of partitions bounded by $O(2^k)$ which is a constant. You do not need the $O(nA)$ dynamic programming algorithm.