This is an odd one but here goes.
First you must recognise that \(\frac{10!}{5!}=\frac{10!*6}{6!}\)
Also
\(n(n-1)(n-2) ... (n-(r-1)) =\frac{n!}{(n-r)!}\)
n and r are both positive integers and \(1\le r\le n\)
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So the question becomes.
Prove \(\frac{n!}{(n-r)!}\) is divisable by r! for all integer values of r from 1 to n inclusive.
Step 1
Prove true for r=1
\(\frac{n!}{(n-1)!}=n\quad and \quad n\div 1! = n\)
Therefore the statement is true for r=1
Step 2
Assume true for r=k
so
\(\frac{n!}{(n-k)!}=Zr!\qquad \text{Z is any integer i.e. }Z\in Z\\ \text{Prove true for }r=k+1\\ \text{that is}\\ \frac{n!}{[n-(k-1)]!}\\=\frac{n![n-(k-1)]}{(n-k)!}\\ =\frac{n!}{(n-k)!}[n-(k-1)]\\ =Zr![n-(k-1)]\\ =Zr! \qquad \text{For a new integer value of Z} \)
So if the expression is a multiple or r! when r=k it will also be a multiple of r! when r=k+1
Step3
Since the expression is a multiple of r! when r=1 it must also be a multiple of r! when r =1, r=3, ...... r=n.
The expression is therefore divisable by r! for all valid values of r
