Given Binomial Expansion
= ( x 5 1 4 + x 5 1 3 ) 60 = {\left( {{{\sqrt x } \over {{5^{{1 \over 4}}}}} + {{\sqrt x } \over {{5^{{1 \over 3}}}}}} \right)^{60}} = ( 5 4 1 x + 5 3 1 x ) 60 ∴ \therefore ∴
T r + 1 = 60 C r . ( x 1 / 2 5 1 / 4 ) 60 − r . ( 5 1 / 2 x 1 / 3 ) r {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} T r + 1 = 60 C r . ( 5 1/4 x 1/2 ) 60 − r . ( x 1/3 5 1/2 ) r
= 60 C r . 5 ( r 4 − 15 + r 2 ) . x ( 30 − r 2 − r 3 ) = {}^{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 ( 4 r − 15 + 2 r ) . x ( 30 − 2 r − 3 r )
= 60 C r . 5 ( 3 r − 60 4 ) . x ( 180 − 5 r 6 ) = {}^{60}{C_r}\,.\,{5^{\left( {{{3r - 60} \over 4}} \right)}}\,.\,{x^{\left( {{{180 - 5r} \over 6}} \right)}} = 60 C r . 5 ( 4 3 r − 60 ) . x ( 6 180 − 5 r ) For x10 term,
180 − 5 r 6 = 10 {{180 - 5r} \over 6} = 10 6 180 − 5 r = 10
⇒ 5 r = 120 \Rightarrow 5r = 120 ⇒ 5 r = 120
⇒ r = 24 \Rightarrow r = 24 ⇒ r = 24 ∴ \therefore ∴
x 10 = 60 C 24 . 5 ( 3 × 24 − 60 4 ) {x^{10}} = {}^{60}{C_{24}}\,.\,{5^{\left( {{{3 \times 24 - 60} \over 4}} \right)}} x 10 = 60 C 24 . 5 ( 4 3 × 24 − 60 )
= 60 C 24 . 5 3 = {}^{60}{C_{24}}\,.\,{5^3} = 60 C 24 . 5 3
= 60 ! 24 ! 36 ! . 5 3 = {{60!} \over {24!\,\,36!}}\,.\,{5^3} = 24 ! 36 ! 60 ! . 5 3 It is given that,
60 ! 24 ! 36 ! . 5 3 = 5 k . l . . . . . . ( 1 ) {{60!} \over {24!\,\,36!}}\,.\,{5^3} = {5^k}\,.\,l \quad ...... (1) 24 ! 36 ! 60 ! . 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 ! = ⌈ n p ⌉ + ⌈ n p 2 ⌉ + ⌈ n p 3 ⌉ + n! = \left\lceil {{n \over p}} \right\rceil + \left\lceil {{n \over {{p^2}}}} \right\rceil + \left\lceil {{n \over {{p^3}}}} \right\rceil + n ! = ⌈ p n ⌉ + ⌈ p 2 n ⌉ + ⌈ p 3 n ⌉ +
∴ \therefore ∴
= ⌈ 60 5 ⌉ + ⌈ 60 5 2 ⌉ + ⌈ 60 5 3 ⌉ + . . . . . = \left\lceil {{{60} \over 5}} \right\rceil + \left\lceil {{{60} \over {{5^2}}}} \right\rceil + \left\lceil {{{60} \over {{5^3}}}} \right\rceil + \ ..... = ⌈ 5 60 ⌉ + ⌈ 5 2 60 ⌉ + ⌈ 5 3 60 ⌉ + .....
= 12 + 2 + 0 + . . . . . = 12 + 2 + 0 + \ ..... = 12 + 2 + 0 + .....
Exponent of 5 in 24!
= ⌈ 24 5 ⌉ + ⌈ 24 5 2 ⌉ + ⌈ 24 5 3 ⌉ + . . . . . . = \left\lceil {{{24} \over 5}} \right\rceil + \left\lceil {{{24} \over {{5^2}}}} \right\rceil + \left\lceil {{{24} \over {{5^3}}}} \right\rceil + \ ...... = ⌈ 5 24 ⌉ + ⌈ 5 2 24 ⌉ + ⌈ 5 3 24 ⌉ + ......
= 4 + 0 + 0 . . . . . . = 4 + 0 + 0 \ ...... = 4 + 0 + 0 ......
Exponent of 5 in 36!
= ⌈ 36 5 ⌉ + ⌈ 36 5 2 ⌉ + ⌈ 36 5 3 ⌉ + . . . . . . . = \left\lceil {{{36} \over 5}} \right\rceil + \left\lceil {{{36} \over {{5^2}}}} \right\rceil + \left\lceil {{{36} \over {{5^3}}}} \right\rceil + \ ....... = ⌈ 5 36 ⌉ + ⌈ 5 2 36 ⌉ + ⌈ 5 3 36 ⌉ + .......
= 7 + 1 + 0 . . . . . . = 7 + 1 + 0 \ ...... = 7 + 1 + 0 ......
∴ \therefore ∴
5 14 5 4 . 5 8 . 5 3 = 5 k {{{5^{14}}} \over {{5^4}\,.\,{5^8}}}\,.\,{5^3} = {5^k} 5 4 . 5 8 5 14 . 5 3 = 5 k
⇒ 5 5 = 5 k \Rightarrow {5^5} = {5^k} ⇒ 5 5 = 5 k
⇒ k = 5 \Rightarrow k = 5 ⇒ k = 5 
