Let $A$ be a non-empty set with $n$ elements.
a) Determine the number of pairs $(X, Y)$ where $X$ is included in $Y$, both included in $A$, and $\text{card}(X) = k$, $\text{card}(Y) = m$; $1 \leq k \leq m \leq n$, with $k$ and $m$ fixed.
b) Show that $\binom{0}{n}\binom{m}{n} + \binom{1}{n}\binom{m-1}{n-1} + \ldots + \binom{m}{n}\binom{0}{n-m} = 2^m \binom{m}{n}$.
My approach: For a, I got $\binom{m}{n}\binom{k}{m}$ using the product rule but I might be wrong. For b) I considered $(X,\,Y)$ with $X\subseteq Y\subseteq A$ where $|X|=k,\,|Y|=m$. The number of such tuples is $\sum_{k=0}^{m} \binom{m}{n} \binom{k}{m} = \binom{m}{n} \left( \sum_{k=0}^{m}\binom{k}{m} \right)$.