Expressing the binomial coefficients in terms of factorials and simplifying algebraically, show that:
(a) $\binom{n}{r}=\frac{n-r+1}{r}\cdot \binom{n}{r-1}$
(b) $\binom{n}{r}=\frac{n}{n-r}\cdot \binom{n-1}{r}$
(c) $n\binom{n-1}{r}=(r+1)\cdot \binom{n}{r+1}$
My attempts:
For (a) (we know by symmetry that $\binom{n}{r} = \binom{n}{r-1}$)
$$\binom{n}{r} = \frac{n-r+1}{r}\cdot \frac{n!}{(r-1)!(n-r)!} = \frac{n!(n-r+1)}{r(r-1)!(n-r)!} = \frac{n!(n-r+1)}{r!(n-r)!}$$
However, I'm unsure of what to do with $n-r+1$, do I divide this from $(n-r)!$ but how as the $+1$ throws me off?
for (b)
$$\binom{n}{r} = \frac{n}{n-r}\cdot \binom{n-1}{r} = \frac{n(n-1)n!}{r!(n+1-r)!}$$
Becuase $(n-r)\times(n-r)! = (n-r+1)!$ But what do I do with $n(n-1)$?
for (c)
$$n\binom{n-1}{r}=(r+1)\cdot \binom{n}{r+1} = \frac{(r+1)n!}{(r+1)!(n-r)!}=\frac{n!}{(r+1-1)!(n-r)!}$$
When I expand the LHS for (c) it looks awfully a lot similar to (b) for example:
$$\frac{n(n-1)n!}{r!(n-(r+1))!}$$
I would deeply appreciate some community support on the right way towards calculating the algebra for these binomial coefficients.
Best Answer
In general, ${n \choose r} \ne {n \choose r-1}$
To solve $(a)$, simply expand LHS.
$\binom{n}{r}=\frac{n!}{r! \cdot (n-r)!} = \frac{n!}{r \cdot (r-1)! \cdot (n-r)!}$
Multiply numerator and denominator by $(n-r+1)$, then,
$\binom{n}{r} = \frac{n-r+1}{r} \cdot \frac{n!}{(r-1)! \cdot (n-r+1)!} = \frac{n-r+1}{r} \cdot {n \choose r-1}$
You should be able to solve $(b)$ and $(c)$ also by expanding LHS and rearranging.