Expressing the binomial coefficients in terms of factorials and simplify

binomial theorembinomial-coefficientsprobability

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.

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.