Given Binomial Expansion

$$ = {\left( {{{\sqrt x } \over {{5^{{1 \over 4}}}}} + {{\sqrt x } \over {{5^{{1 \over 3}}}}}} \right)^{60}}$$

$$\therefore$$ General term

$${T_{r + 1}} = {}^{60}{C_r}\,.\,{\left( {{{{x^{1/2}}} \over {{5^{1/4}}}}} \right)^{60 - r}}\,.\,{\left( {{{{5^{1/2}}} \over {{x^{1/3}}}}} \right)^r}$$

$$ = {}^{60}{C_r}\,.\,{5^{\left( {{r \over 4} - 15 + {r \over 2}} \right)}}\,.\,{x^{\left( {30 - {r \over 2} - {r \over 3}} \right)}}$$

$$ = {}^{60}{C_r}\,.\,{5^{\left( {{{3r - 60} \over 4}} \right)}}\,.\,{x^{\left( {{{180 - 5r} \over 6}} \right)}}$$

For x¹⁰ term,

$${{180 - 5r} \over 6} = 10$$

$$ \Rightarrow 5r = 120$$

$$ \Rightarrow r = 24$$

$$\therefore$$ Coefficient of $${x^{10}} = {}^{60}{C_{24}}\,.\,{5^{\left( {{{3 \times 24 - 60} \over 4}} \right)}}$$

$$ = {}^{60}{C_{24}}\,.\,{5^3}$$

$$ = {{60!} \over {24!\,\,36!}}\,.\,{5^3}$$

It is given that,

$${{60!} \over {24!\,\,36!}}\,.\,{5^3} = {5^k}\,.\,l$$ ...... (1)

Also given that, l is coprime to 5 means l can't be multiple of 5. So we have to find all the factors of 5 in 60!, 24! and 36!

[Note : Formula for exponent or degree of prime number in n!.

Exponent of p in $$n! = \left\lceil {{n \over p}} \right\rceil + \left\lceil {{n \over {{p^2}}}} \right\rceil + \left\lceil {{n \over {{p^3}}}} \right\rceil + $$ ..... until 0 comes

here p is a prime number. ]

$$\therefore$$ Exponent of 5 in 60!

$$= \left\lceil {{{60} \over 5}} \right\rceil + \left\lceil {{{60} \over {{5^2}}}} \right\rceil + \left\lceil {{{60} \over {{5^3}}}} \right\rceil + $$ .....

$$ = 12 + 2 + 0 + $$ .....

$$ = 14$$

Exponent of 5 in 24!

$$ = \left\lceil {{{24} \over 5}} \right\rceil + \left\lceil {{{24} \over {{5^2}}}} \right\rceil + \left\lceil {{{24} \over {{5^3}}}} \right\rceil + $$ ......

$$ = 4 + 0 + 0$$ ......

$$ = 4$$

Exponent of 5 in 36!

$$ = \left\lceil {{{36} \over 5}} \right\rceil + \left\lceil {{{36} \over {{5^2}}}} \right\rceil + \left\lceil {{{36} \over {{5^3}}}} \right\rceil + $$ .......

$$ = 7 + 1 + 0$$ ......

$$ = 8$$

$$\therefore$$ From equation (1), exponent of 5 overall

$${{{5^{14}}} \over {{5^4}\,.\,{5^8}}}\,.\,{5^3} = {5^k}$$

$$ \Rightarrow {5^5} = {5^k}$$

$$ \Rightarrow k = 5$$