The number of rational terms in the expansion of $(1+\sqrt{2}+\sqrt[3]{5}{)}^6$ is

A term in the expansion  of $(1+\sqrt{2}+\sqrt[3]{5}{)}^6$ is of the form

$\frac{6!}{r!s!(6-r-s)!}\left(1{)}^{6-r-s}\right(\sqrt{2}{)}^r(\sqrt[3]{5}{)}^s$

$=\frac{6!}{r!s!(6-r-s)!}\left(2^{r/2}\right)\left(5^{s/3}\right)$

This term will be rational if $2^{r/2}$,and $5^{s/3}$ are both rational
numbers. This is possible if and only if r is a multiple of 2
and s is a multiple of 3. Possible values of r are 0, 2, 4 and
6 whereas the possible values of s are 0, 3 and 6. Also note that

For $0\le r+s\le 6$. s can take value 0, 3 or 6.

For $r=0$ s can take value 0 or 3.

For r = 4, s can take value 0.

For r = 6, s can take value 0.

Thus, the number of rational terms is 3 + 2 + 1 + 1 = 7