First of all, notice that the picture is a little misleading; it’s always enough to check configurations where the graph induced by the C-molecules is a chain. It’s a little tough to formally prove, but the rough idea is that this setup maximizes both the number of I-molecules that you can attach and the number of H-molecules that you can further attach (for a fixed number of I).

Another important observation is that there should be an even number of H-molecules (the sum of degrees in the graph is even).

Then, let’s fix the number $$$C$$$ of C ($$$N \geq 30$$$ implies $$$C \geq 2$$$) molecules. For each fixed $$$C$$$, you have two cases:

1. $$$H = 0$$$: The constraints forbid that, so we can safely ignore this case.
2. $$$H = 2 + 2H’$$$: Put two of the molecules in the endpoints. Now you have $$$C$$$ interchangeable C-molecules each of which can support either one I-molecule or one or two H-molecules. It turns out that you can attach the remaining molecules iff $$$I + 2H’ \leq C$$$. So you just need to count the number of pairs $$$(I, H’)$$$ with $$$3I + 2H’ = N - 5C - 2$$$ and $$$H’, I \geq 0$$$ and $$$I \leq C-2H’$$$. Again, see if you can derive a formula for computing that.