Ferrers graph question. Use Ferrers
graphs to show that the number of partitions of n is equal to the number of partitions of 2n that have exactly n summands.
Now i drew this out for $n=1, n=2, n=3$ and it appears to be true i also (rather fruitlessly) tried to show this was the case by induction.
I have noticed several related questions regarding odd parts and distinct parts being the number ( which we proved in class) i feel like this is the same question if i can figure out how to re word it what i am asked to show would follow from it almost immediately. i would like to avoid generating functions for this result if possible. 
 A: The following example should make clear the bijection between integer partitions of $2n$ into exactly $n$ parts and integer partitions of $n$. 
In this case $n=8$ 
$$\begin{array}{cccccccc}
\bbox[black,5px]{\phantom{H}} & \bbox[white,5px]{\phantom{H}} & \bbox[white,5px]{\phantom{H}} & \bbox[white,5px]{\phantom{H}} & \bbox[white,5px]{\phantom{H}} & \bbox[white,5px]{\phantom{H}} & \bbox[white,5px]{\phantom{H}} & \bbox[white,5px]{\phantom{H}} \\
\bbox[black,5px]{\phantom{H}} & \bbox[black,5px]{\phantom{H}} & \bbox[black,5px]{\phantom{H}} & \bbox[white,5px]{\phantom{H}} & \bbox[white,5px]{\phantom{H}} & \bbox[white,5px]{\phantom{H}} & \bbox[white,5px]{\phantom{H}} & \bbox[white,5px]{\phantom{H}} \\
\bbox[black,5px]{\phantom{H}} & \bbox[black,5px]{\phantom{H}} & \bbox[black,5px]{\phantom{H}} & \bbox[black,5px]{\phantom{H}} & \bbox[white,5px]{\phantom{H}} & \bbox[white,5px]{\phantom{H}} & \bbox[white,5px]{\phantom{H}} & \bbox[white,5px]{\phantom{H}} \\
\bbox[blue,5px]{\phantom{H}} & \bbox[blue,5px]{\phantom{H}} & \bbox[blue,5px]{\phantom{H}} & \bbox[blue,5px]{\phantom{H}} & \bbox[blue,5px]{\phantom{H}} & \bbox[blue,5px]{\phantom{H}} & \bbox[blue,5px]{\phantom{H}} & \bbox[blue,5px]{\phantom{H}} \\
\hline 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8
\end{array}$$
we see that any partition of $2n$ into $n$ parts has exactly $n$ blocks in the bottom row. Deleting this row for every partition into exactly $n$ parts yields a unique partition of $n$. 
Conversely, any partition of $n$ can be turned into one of $2n$ into $n$ parts by adding a full bottom row.
symbolically we write total partitions of $a$ as $p(a)$ and of $a$ into exactly $b$ parts as $p(a,b)$, then:
$$p(2n,n)=p(n)=\sum_{k=1}^{n}p(n,k)$$
