Is 100!/(50! x 2^50) an integer?

I just read this puzzle somewhere and realized that this is similar to my earlier post How many zeros does 1000! end with?

Here is the solution to this puzzle,

There is no doubt that 100!/50! is an integer. Hence we just need to find whether 100 x 99 x 98 x ... x 51 will be divisible by 2^50 or not. The mathematical theorem discussed in my earlier post helps us to solve this puzzle as well

let us find the exponents of 2 in 100!

   [100/2] + [100/2^2 ] + [100/2^3 ] + [100/2^4 ] + [100/2^5 ] + [100/2^6 ]

      = 50 + 25 + 12 + 6 +3+1 = 97

Therefore 100! is multiples of 2^97

let us find the exponents of 2 in 50!

    [50/2] + [50/2^2 ] + [50/2^3 ] + [50/2^4 ] + [50/2^5 ]

       = 25 + 12 + 6 +3+1 = 47

Therefore 50! is multiples of 2^47

From the above it is evident that 100! / 50! is multiples of 2^50. Hence 100!/(50! x 2^50) is an integer