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

We have,

$(1+\sqrt{2}+\sqrt[3]{3}{)}^6=\sum _{r+s+t=6} \frac{6!}{r!s!t!}\times 2^{s/2}{3}^{t/3}$

Now, $\frac{6!}{r!s!t!}\times 2^{s/2}\times 3^{t/3}$ will be an integer, if $s$ is a multiple of

$2$and $t$is a multiple of 3 i.e..

$s=0, 2, 4, 6$ and $t=0, 3, 6$

But, $r+s+t=6$  Therefore, we have the following 

possibilities.

$\begin{array}{r}r=6,s=0,t=0\\ r=3,s=0,t=3\\ r=0,s=0,t=6\\ r=4,s=2,t=0\\ r=1,s=2,t=3\\ r=2,s=4,t=0\\ r=0,s=6,t=0\end{array}$

Hence, there are $7$rational terms in the expansion of

$(1+\sqrt{2}+\sqrt[3]{3}{)}^6$