Calculating number of unique combinations for this scenario.

First, recall that the way to partition a set of $n$ elements in $k$ (possibly empty) ordered parts, is $\binom{n+k-1}{k-1}$. The set is going to be given by the choices of the picked items, the number of parts by the number of possible choices (e.g. we associate the choice C1, C1, C3, C4 to the ordered set partition of $5$ given by $(2, 0, 1, 1, 0)$, as we choose C1 two times, C2 zero times, C3 one time, C4 one time, C5 zero times).

In set A, you have to choose one item out of one, so $n=k=1$ and you have $1$ choice.

In set B, you have to choose four items out of two, so $n=4$, $k=2$, and you have $\binom{4+2-1}{2-1} = 5$ choices. (As additional check, it is easy to see that you can choose B1 four times, B1 three times and B2 once, B1 and B2 both twice, B1 once and B2 three times, or B2 four times, so $5$ choices)

In set C, you have to choose four items out of five, so $n=4$, $k=5$, and you have $\binom{4+5-1}{5-1} = 70$ choices.

In total, the number of unique combinations is \[ 1 \cdot 5 \cdot 70 = 350 \]