Let $Q(n, k)$ be the number of partitions of $n$ into $k$ distinct, unequal parts. Prove $Q(n + {k + 1\choose 2}, k)$ is equal to the number of ways to partition $n$ into at most $k$ parts (parts can be equal here).
I am not so sure how to approach this problem because the number $n + {k+1\choose 2}$ seems arbitrary. I would guess that we need to use the fact that no parts are equal for $Q(n + {k+1\choose 2}, k)$, but I have already tried for many hours with no luck. I would appreciate any help.
Something I noticed that may or may not help is ${k+1\choose 2}$ is the sum of all positive integers up to $k$.
Best Answer
Is there any bijective map between these two kind of partitions?
Hint. Note that if $0\leq x_1\leq x_2\leq \dots \leq x_k$ with $$x_1+x_2+\dots +x_k=n$$ then $1\leq y_1<y_2< \dots <y_k$ with $$y_1+y_2+\dots +y_k=n+1+2+\dots+k=n+{k+1\choose 2}$$ where $y_i=x_i+i$ for $i=1,2,\dots,k$, and, as you noticed ${k+1\choose 2}$ is the sum of all positive integers up to $k$.