find the number of terms free from radical sign in the expansion of { 1+3^1/3+7^1/7}^10

Dear student,
as we know that general term of ${\left(a_1+a_2+a_3+.........+a_k\right)}^n$ is $\frac{n!}{r_1! r_2! r_3!......r_k!}{a_1}^{r_1}{\left(a_2\right)}^{r_2}{\left(a_3\right)}^{r_3}.......{\left(a_k\right)}^{r_k}$
now general term of ${\left(1+3^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}+7^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$7$}\right.}\right)}^{10}$ will be $\frac{10!}{a!.b!.c!}{\left(1\right)}^a{\left(3^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\right)}^b{\left(7^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$7$}\right.}\right)}^c$
now term will be free from radical sign if 
b is a multiple of 3
and c is a multiple of 7 and we need a+b+c=10
so possible combination are 
$\begin{array}{ccc}a& b& c\\ 10& 0& 0\\ 7& 3& 0\\ 4& 6& 0\\ 3& 0& 7\\ 0& 3& 7\end{array}$
clearly there are total 5 terms free from radical sign

enjoy