Question 858584
You do not need to calculate a number as astronomical as 5000!,
because {{{5000!/(5000-50)!=5000!/4950!=5000*4999*4998*"..."*4953*4952*4951}}} ,
and that is slightly less astronomical.
According to the combinations function in the Excel spreadsheet program in my computer, 5000C50 is {{{2.2839*10^120}}} .
That must also an estimate, because there is no way it could calculate all 120 decimal places.
I also calculated 5000C50 as
{{{5000*4999*4998*"..."*4953*4952*4951/50!=6.9432*10^184/(3.04141*10^64)}}}
Excel also calculated {{{5000*4999*4998*"..."*4953*4952*4951=6.9432*10^184}}} for me.
I do not know if you can do that with a calculator.
A crude approximation of {{{5000!/4950!=5000*4999*4998*"..."*4953*4952*4951}}}
would be {{{5000^50=(5*1000)^50=(5*10^3)^50=5^50*10^150=8.8178*10^34*10^150=8.8178*10^184}}}.
A better approximation for
{{{5000*(4999*4951)*(4998*1952)*"..."*(4976*4974)*4975=5000*(4975^2-24^2)*(4975^2-23^2)*"..."*(4975^2-1^2)*4975}}}
would be {{{5000*4975^49=6.9475*10^184}}} .