If the coefficient of x¹⁰ in the binomial expansion of $${\left( {{{\sqrt x } \over {{5^{{1 \over 4}}}}} + {{\sqrt 5 } \over {{x^{{1 \over 3}}}}}} \right)^{60}}$$ is ${5^k}\,.\,l$, where l, k $\in$ N and l is co-prime to 5, then k is equal to _____________.

<p>Given Binomial Expansion</p> <p>$$ = {\left( {{{\sqrt x } \over {{5^{{1 \over 4}}}}} + {{\sqrt x } \over {{5^{{1 \over 3}}}}}} \right)^{60}}$$</p> <p>$\therefore$ General term</p> <p>$${T_{r + 1}} = {}^{60}{C_r}\,.\,{\left( {{{{x^{1/2}}} \over {{5^{1/4}}}}} \right)^{60 - r}}\,.\,{\left( {{{{5^{1/2}}} \over {{x^{1/3}}}}} \right)^r}$$</p> <p>$$ = {}^{60}{C_r}\,.\,{5^{\left( {{r \over 4} - 15 + {r \over 2}} \right)}}\,.\,{x^{\left( {30 - {r \over 2} - {r \over 3}} \right)}}$$</p> <p>$$ = {}^{60}{C_r}\,.\,{5^{\left( {{{3r - 60} \over 4}} \right)}}\,.\,{x^{\left( {{{180 - 5r} \over 6}} \right)}}$$</p> <p>For x¹⁰ term,</p> <p>${{180 - 5r} \over 6} = 10$</p> <p>$\Rightarrow 5r = 120$</p> <p>$\Rightarrow r = 24$</p> <p>$\therefore$ Coefficient of $${x^{10}} = {}^{60}{C_{24}}\,.\,{5^{\left( {{{3 \times 24 - 60} \over 4}} \right)}}$$</p> <p>$= {}^{60}{C_{24}}\,.\,{5^3}$</p> <p>$= {{60!} \over {24!\,\,36!}}\,.\,{5^3}$</p> <p>It is given that,</p> <p>${{60!} \over {24!\,\,36!}}\,.\,{5^3} = {5^k}\,.\,l$ ...... (1)</p> <p>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!</p> <p>[Note : Formula for exponent or degree of prime number in n!.</p> <p>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</p> <p>here p is a prime number. ]</p> <p>$\therefore$ Exponent of 5 in 60!</p> <p>$$= \left\lceil {{{60} \over 5}} \right\rceil + \left\lceil {{{60} \over {{5^2}}}} \right\rceil + \left\lceil {{{60} \over {{5^3}}}} \right\rceil + $$ .....</p> <p>$= 12 + 2 + 0 +$ .....</p> <p>$= 14$</p> <p>Exponent of 5 in 24!</p> <p>$$ = \left\lceil {{{24} \over 5}} \right\rceil + \left\lceil {{{24} \over {{5^2}}}} \right\rceil + \left\lceil {{{24} \over {{5^3}}}} \right\rceil + $$ ......</p> <p>$= 4 + 0 + 0$ ......</p> <p>$= 4$</p> <p>Exponent of 5 in 36!</p> <p>$$ = \left\lceil {{{36} \over 5}} \right\rceil + \left\lceil {{{36} \over {{5^2}}}} \right\rceil + \left\lceil {{{36} \over {{5^3}}}} \right\rceil + $$ .......</p> <p>$= 7 + 1 + 0$ ......</p> <p>$= 8$</p> <p>$\therefore$ From equation (1), exponent of 5 overall</p> <p>${{{5^{14}}} \over {{5^4}\,.\,{5^8}}}\,.\,{5^3} = {5^k}$</p> <p>$\Rightarrow {5^5} = {5^k}$</p> <p>$\Rightarrow k = 5$</p>