1. General term of the expansion

For $\left(4^{1/4}+5^{1/6}\right)^{120},$ the general term is $T_{r+1}=\binom{120}{r}\left(4^{1/4}\right)^{120-r}\left(5^{1/6}\right)^r, \qquad r=0,1,2,\dots,120.$

2. Simplify the powers

Since $4^{1/4}=(2^2)^{1/4}=2^{1/2},$ we get $\left(4^{1/4}\right)^{120-r}=4^{(120-r)/4}=2^{(120-r)/2}.$ Also, $\left(5^{1/6}\right)^r=5^{r/6}.$

So the general term becomes $T_{r+1}=\binom{120}{r} \cdot 2^{(120-r)/2} \cdot 5^{r/6}.$

3. Condition for a rational term

A term is rational if both exponents are integers:

• $\dfrac{120-r}{2}$ must be an integer,
• $\dfrac{r}{6}$ must be an integer.

Now, if $r$ is divisible by $6$, then it is automatically even, so both conditions are satisfied.

Hence the required condition is: $r\equiv 0 \pmod{6}.$

4. Count such values of $r$

Since $r$ ranges from $0$ to $120$, the possible values are $r=0,6,12,\dots,120.$

This is an arithmetic progression with:

• first term $0$,
• last term $120$,
• common difference $6$.

Number of terms: $\frac{120-0}{6}+1=20+1=21.$

5. Final answer

Therefore, the number of rational terms is $\boxed{21}.$