The number of rational terms in the expansion of (2^1/3+3^1/2)^60 is

The general term is $T_{r+1}=\binom{60}{r}\,2^{(60-r)/3}\,3^{r/2}$, with $0\le r\le60$. It is rational exactly when both exponents are integers: $\frac{60-r}{3}$ is an integer iff $r$ is a multiple of $3$ (since $60$ is), and $\frac r2$ is an integer iff $r$ is even. So $r$ must be a multiple of $6$: $r=0,6,12,\dots,60$. That gives $\frac{60}{6}+1=11$ rational terms. Counting only multiples of $3$ would give $21$ and only evens $31$, so the common condition matters.