Proving $\binom{m}{n} + \binom{m}{n-1} = \binom{m+1}{n}$ algebraically

Question

I am working through the exercises and have spent half a day on one problem so I decided to get some help because I can't figure it out.

Show that if $n$ is a positive integer at most equal to $m$, then $$ \binom{m}{n} + \binom{m}{n-1} = \binom{m+1}{n} $$ I expand the sum of two binomial coefficients and get $$ \frac{m!}{n!(m-n)!} + \frac{m!}{(n-1)!(m-n+1)!} $$ Now, I have trouble summing up the two fractions, specifically, finding the common denominator of two fractions. I tried all sorts of things I could think of but can't seem to get to $n!(m-n+n)!$ which is in the solutions at the back of the book.

My last try $$ \frac{m!(n!(m-n)!) + m!((n-1)!(m-n+1)!)}{n!(m-n)!(m-n+1)!(n-1)!} = $$ $$ = \frac{m!(n!(m-n)! + (n-1)!(m-n+1)!)}{n!(m-n+1)!(m-n)!(n-1)!} = $$ $$ = \frac{m!(mn!-nn!+((mn-n^2+n)!-(m-n+1)!)!}{n!(m-n+1)!(m-n)!(n-1)!} $$

And so on, I feel like I am missing something as I don't see any opportunity to simplify this mess.

Answer 1

Lazy people like me should do it as follows. Define $R(m,n)$ as:

$$R(m,n) = \frac{\binom{m}{n}+\binom{m}{n-1}}{\binom{m+1}{n}}$$

It should be clear that $R(m,n)$ is some rational function and we want to prove that this is actually identical to 1. Now, because it is manifestly a rational function, the algebra is guaranteed to simplify a lot making this method suitable for lazy people. We have:

$$\begin{split} \frac{\binom{m}{n}}{\binom{m+1}{n}} &= \frac{m-n+1}{m+1}\\[1em] \frac{\binom{m}{n-1}}{\binom{m+1}{n}} &= \frac{n}{m+1} \end{split} $$

Clearly $R(m,n)$ is indeed 1.

Answer 2

If you really want to prove this identity using bloody/gory algebraic manipulations, then note the following: \begin{align} \binom{m}{n}+\binom{m}{n-1} &= \frac{m!}{n!(m-n)!}+\frac{m!}{(n-1)!(m-n+1)!}\\[1em] &= \frac{m!}{n!(m-n)!}+\frac{n\cdot m!}{n!(m-n)!(m-n+1)}\\[1em] &= \frac{m!(m-n+1)+n\cdot m!}{n!(m-n)!(m-n+1)}\\[1em] &= \frac{m!(m-n+1+n)}{n!(m-n)!(m-n+1)}\\[1em] &= \frac{m!(m+1)}{n!(m-n)!(m-n+1)}\\[1em] &= \frac{(m+1)!}{n!(m+1-n)!}\\[1em] &= \binom{m+1}{n}. \end{align}

Answer 3

There is a very simple combinatorics proof for that.

Assume that we have $m+1$ objects called $a_0, \dots, a_m$ and we want to select $n$ objects out of them. There are $\binom {m+1}{n}$ ways to do this.

This selection can be done in two ways. Either we include the first element in our selection or not. If we include the first element we are left with selecting $n-1$ objects out of the remaining $m$ elements that can be done in $\binom{m}{n-1}$ ways. If we do not include the first element we need to select $n$ objects out of the remaining $m$ elements that can be done in $\binom {m}{n}$ ways. Therefore we have,

$$ \binom {m+1}{n} = \binom {m}{n-1} + \binom{m}{n} $$

Answer 4

The LCM of $n!(m-n)!$ and $(n-1)!(m-n+1)!$ is $n!(m-n+1)!$. This must be the case because $n!=(n-1)!\cdot n$ and $(m-n+1)!=(m-n)!\cdot (m-n+1)$ . With this in mind, you should be able to easily convert the two fractions into a common denominator. Namely,

$$\frac{m!\cdot(m-n+1)}{n!(m-n+1)!}+\frac{m!\cdot n}{n!(m-n+1)!}$$

It's really not as complicated as it looks, once you have the common denominator correct.

Answer 5

Proof using formula: $$ {{m\choose n} := {m(m-1)\ldots(m-n+1)\over n!}} $$

$$ \eqalign{{m\choose n}+{m\choose n-1}&= {m(m-1)\ldots(m-n+1)\over n!}+{m(m-1)\ldots(m-(n-1)+1)\over (n-1)!}\cr &={m(m-1)\ldots(m-n+2)\cdot(m-n+1)\over n!}+{m(m-1)\ldots(m-n+2)\over (n-1)!} \cdot {n \over n}\cr &={m(m-1)\ldots(m-n+2)\cdot[(m-n+1)+n]\over n!}\cr &={m(m-1)\ldots(m-n+2)(m+1)\over n!}\cr &={(m+1)m(m-1)\ldots[(m+1)-n+1)]\over n!}\cr &={m+1\choose n}. } $$