Rational terms of expansion

After expansion the general term is,

$Constant \times (2^{\frac{1}{2}})^x \times (3^{\frac{1}{3}})^y \times (5^{\frac{1}{6}} )^z$

Where , $x+y+z=10 \dots \dots \dots \dots(1)$

A term becomes rational iff $x\in\{0,2,4,6,8,10\} ; \ y\in\{0,3,6,9\} ; \ z\in\{0,6\}$

It can be easily found that only $(10,0,0) , (4,6,0) , (4,0,6)$ these three triples make $(1)$ true.

So there are three rational term in the expansion of $\left( 2^{\frac{1}{2}}+3^{\frac{1}{3}}+5^{\frac{1}{6}} \right)^{10}$

And they are

${10\choose10}(2^{\frac{1}{2}})^{10} \times (3^{\frac{1}{3}})^0 \times (5^{\frac{1}{6}} )^0=32$

${10\choose4}(2^{\frac{1}{2}})^{4} \times (3^{\frac{1}{3}})^6 \times (5^{\frac{1}{6}} )^0=7560$

${10\choose4}(2^{\frac{1}{2}})^{4} \times (3^{\frac{1}{3}})^0 \times (5^{\frac{1}{6}} )^6=4200$

So the answer is $32+7560+4200=\fbox{11792}$