Question

please help to prove by a combinatorial argument that rC(n,r) = nC(n-1,r-1) for 1≤r≤n

Answer 1

It is probably easiest to see this relation if you expand, in factorial notation, the right side of the equation;

\[ n{n-1 \choose r-1}=n\frac{(n-1)!}{(r-1)!(n-1-(r-1))!}=n\frac{(n-1)!}{(r-1)!(n-r)!}\]

Now the left side:

\[ r{n\choose r }=r\frac{n!}{r!(n-r)!}\]

Notice how the left side and right side expression are very similar. To complete the proof:
Notice that \(n!=n(n-1)!\) and \(r!=r(r-1)!\). use this for the left side expression.