Number of ways constructing a $k$-clique. I have a complete graph $A=K_{n}$ and a natural numbers $m,k$

Determine the number of all subgraphs in $A$ that have exactly $m$ edges and at least one $k$-clique. The same with subgraphs that don't contain any $k$-clique. Let's denote these numbers respectively as $A(m),B(m)$ 

I would like to focus on following case:
$${n\choose 2}\ge m \ge 0$$
Is this true that $$A(m)+B(m)={{n\choose 2}\choose m}$$?
 A: Since the number of all subgraphs in $A$ that have exactly $m$ edges and at least one $k$-clique is complimentary to the number of subgraphs that have exactly $m$ edges and don't contain any $k$-clique, $A(m) + B(m)$ is just the number of all subgraphs with $m$ edges. So choose $m$ edges from $n \choose 2$ edges: $$A(m) + B(m) = {{n\choose 2}\choose m}.$$
In the comments it was shown that the number of all subgraphs in $A$ that have exactly $m$ edges and at least one $k$-clique is $$ {n \choose k} \cdot {{{n \choose 2} - {k \choose 2}} \choose {m - {k \choose 2}}}. $$
Therefore the number of subgraphs that have exactly $m$ edges and don't contain any $k$-clique is
$$ {{n\choose 2}\choose m} - {n \choose k} \cdot {{{n \choose 2} - {k \choose 2}} \choose {m - {k \choose 2}}}.$$

For completeness, I just repeat the proof from the comments (due to NikoWielopolski) showing how to calculate the number of all subgraphs in $A$ that have exactly $m$ edges and at least one $k$-clique.
Start by selecting $k$ vertices from $A$ (together with all edges between these vertices - this will then be a $k$-clique) - to do this we choose $k$ vertices from $n$: so there are $n \choose k$ ways to do this. The $k$-clique has $k \choose 2$ edges, so we need to add $m - {k \choose 2}$ edges outside of the clique to complete the subgraph. The total amount of edges outside of the $k$-clique we have chosen is ${n \choose 2} - {k \choose 2}$, so there are $${{{n \choose 2} - {k \choose 2}} \choose {m - {k \choose 2}}}$$ ways to do this.
Combining the choice of vertices for the $k$-clique with the choice of edges outside the clique yields the total number of subgraphs with exactly $m$ edges and at least one $k$-clique: $$ {n \choose k} \cdot {{{n \choose 2} - {k \choose 2}} \choose {m - {k \choose 2}}}. $$
