# show number of partitions of $n$ equals to number of partitions of $n-k$

Let $$n,k\in\mathbb{Z}^+$$ and $$n\geq k$$. Suppose $$n = \lambda_1 + \lambda_2 + \cdots + \lambda_k$$ is an integer partition of $$n$$, and $$\lambda_1 \geq\lambda_2\geq\cdots\geq\lambda_k$$.

Show (the number of partitions of $$n$$ in which $$\lambda_1=k$$) = (the number of partitions of $$n-k$$ in which $$\mu_i\leq k,\forall i$$)

So I think first to derive the generating functions for both sides and then prove they're equal to each other? And the generating functions of $$p(n)$$ is $$P(n)=\sum\limits_{n=0}^\infty p_nx^n=\prod\limits_{i=1}^\infty \frac{1}{1-x^i}$$?

• This is muddled: you are using the function symbol $p$ for two different things. I can't quite work out what you're asking, but I suspect Ferrers diagrams is the tool to use. – Lord Shark the Unknown Nov 1 '18 at 16:36
• When you have your brackets "(the number of...", do you mean the cardinality of the set described by the stuff in the brackets multiplied by what the bracket is next to? – Sam Streeter Nov 1 '18 at 16:38
• @SamStreeter I made some edits does that help? – Thomas Nov 1 '18 at 16:41

There's an obvious bijection $$\{\lambda_1 = k,\lambda_2,\ldots,\lambda_k\} \leftrightarrow \{\mu_1,\ldots,\mu_{k-1}\} = \{\lambda_2,\ldots,\lambda_k\}$$ between partitions of $$n$$ into $$k$$ elements $$\lambda_i$$, $$i=1,\ldots,k$$ with $$\lambda_1 = k$$, $$\lambda_1 \geq \ldots \geq \lambda_k$$ and partitions of $$n-k$$ into $$k-1$$ elements $$\mu_j$$, $$j=1,\ldots,{k-1}$$ with all elements less than or equal to $$k$$ AND $$\mu_1 \geq \mu_2 \geq \ldots \mu_{k-1}$$. Indeed, $$k = \lambda_1 \geq \lambda_i$$ for all $$\lambda_i$$ with $$i \geq 2$$ and $$\lambda_2 + \ldots + \lambda_k = n - k$$.