Help understand second step in proof of ${n\choose m-1}+{n\choose m}={n+1\choose m}$

Question

I have problem understanding second step in this proof: \begin{align} \binom n{m-1}+\binom nm&=\frac{n!}{(m-1)!(n-m+1)!}+\frac{n!}{m!(n-m)!}\\ &=\frac{n!m+n!(n-m+1)}{m!(n-m+1)!}\\ &=\frac{n!(n-m++1+m)}{m!(n-m+1)!}\\ &=\frac{n!(n+1)}{m!(n-m+1)!}\\ &=\frac{(n+1)!}{m!(n-m+1)!}=\binom{n+1}m \end{align}

Why is it not like this: \begin{align} &=\frac{n!m!(n-m)!+n!(m-1)!(n-m+1)!}{(m-1)!(n-m+1)!m!(n-m)!}\\ \end{align} But like this: \begin{align} &=\frac{n!m+n!(n-m+1)}{m!(n-m+1)!}\\ \end{align} Does it just skip some step/steps that are clear to people with more mature math skills? I am doing exercises from Serge Lang's Basic mathematics as self study. Proof in question is from here: Binomial coefficient proof for ${n\choose m-1}+{n\choose m}={n+1\choose m}$

Answer 1

The skipped step looks as following $$ \begin{align} \binom n{m-1}+\binom nm&=\frac{n!}{(m-1)!(n-m+1)!}+\frac{n!}{m!(n-m)!}\\ &{=\frac{n!\color{red}m}{(m-1)!\color{red}m(n-m+1)!}+\frac{n!\color{red}{(n-m+1)}}{m!(n-m)!\color{red}{(n-m+1)}}}\\ &=\frac{n!m}{m!(n-m+1)!}+\frac{n!(n-m+1)}{m!(n-m+1)!}.\\ \end{align} $$

Answer 2

They do skip an intermediate step or two maybe. You just have to note that $m!=m(m-1)!$ and similarly $(n-m+1)! = (n-m+1)(n-m)!$. This allows you rewrite both terms on the right in the first line to expressions which have $m!(n-m+1)!$ in the denominator.

Answer 3

Using

 m!=m(m-1)!
(n-m+1)!=(n-m+1)!(n-m)

In step 3 on first term I multiple and divide by m! and in second term I multiply and divide by (n-m+1)!

In Step 4 I have opened factorial as I mentioned in using part and the rest is simple LCM now.

\begin{align} \binom n{m-1}+\binom nm&=\frac{n!}{(m-1)!(n-m+1)!}+\frac{n!}{m!(n-m)!}\\ &=\frac{m!n!}{m!(m-1)!(n-m+1)!}+\frac{(n-m+1)!n!}{(n-m+1)!m!(n-m)!}\\ &=\frac{m(m-1)!n!}{m!(m-1)!(n-m+1)!}+\frac{(n-m+1)(n-m)!n!}{(n-m+1)!m!(n-m)!}\\ &=\frac{mn!}{m!(n-m+1)!}+\frac{(n-m+1)n!}{(n-m+1)!m!}\\ &=\frac{n!m+n!(n-m+1)}{m!(n-m+1)!}\\ &=\frac{n!(n-m++1+m)}{m!(n-m+1)!}\\ &=\frac{n!(n+1)}{m!(n-m+1)!}\\ &=\frac{(n+1)!}{m!(n-m+1)!}=\binom{n+1}m \end{align}

Answer 4

A proof using combinatory (more elegant).

Let consider a group $n+1$ people. Then, $\binom{n+1}{m}$ is the number of groups of $m$ people that we can form. Let Frank a member of these $n+1$ people. Notice that to make a group with $m$ people, we can consider group where Frank belong (and there are $\binom{n}{m-1}$) and Group where Frank do not belong (and there are $\binom{n}{m}$). Finally, $$\binom{n+1}{m}=\binom{n}{m-1}+\binom{n}{m}.$$

Answer 5

Just multiply the three terms in the formulas for $\binom{n+1}{m}, \binom{n}{m-1}$ and $\binom{n}{m}$ with $\frac{(n - m + 1)!m!}{n!}$.
Then you get: $\frac{m!}{(m - 1)!} + \frac{(n - m + 1)!}{(n - m)!} = \frac{(n + 1)!}{n!}$.
And, after simplifying: $(m) + (n - m + 1) = (n + 1)$, which is true.