Hello, musngiburger!
Find the number of trailing zeros in the ordinary decimal representation of the number: .$\displaystyle {123\choose45}$
We have: .$\displaystyle {123\choose45} \:=\:\frac{123!}{45!\,78!} $
The numerator is: .$\displaystyle 79\cdot80\cdot81\,\cdots\, 123$
The denominator is: .$\displaystyle 1\cdot2\cdot3\,\cdots\,45$
The number of trailing zeros depends on the number of factors-of-5
. . that remain in the numerator after reducing the fraction.
The numbers in the numerator which have a factor-of-5 are:
. . $\displaystyle 80,85,90,95,100,105,110,115,120$
There are ten factors-of-5. .(100 has two.)
The numbers in the denominator which have a factor of 5 are:
. . $\displaystyle 5,10,15,20,25,30,35,40,45$
There are ten factors-of-5. .(25 has two.)
We find that all the factors-of-5 cancel out.
Therefore, there are no trailing zeros in $\displaystyle {123\choose45}$