Question: Question B-2 of the [2000 Putnam Exam](http://kskedlaya.org/putnam-archive/2000.pdf) was answered [here](https://math.stackexchange.com/questions/1165229/how-can-we-prove-that-a-binomial-coefficient-n-choose-mis-divisible-by-the-rat). in all the answers everyone assumes that $\binom{m}{n}$ is an integer for $m \geq n \geq 1$ integers, but gives no further explanation as to why.

When I tried to prove it myself, I was met with complicated expressions that prevented me from continuing.
My question is: what is the simple way to prove that $\binom{m}{n}$ is an integer for $m \geq n$ positive integers?

Answer: 
Newton's symbol is used to compute the number of ways to choose n elements from a set with m elements. It is assumed that the order of the chosen elements is irrelevant, that is, {a, b} = {b, a}. That is regarded as the same choice.

So you look at m elements and take n of them. How many ways are there to do that if order is to be accounted for? You have m options for the first, (m-1) options for the second etc, all the way up to the last element: (m-(n-1)) options. Hence there are $\frac{m!}{(m-n)!}$ choices.

But remember, one ought to discard all the unnecessary permutations! Since n elements have been chosen, one should divide the outcome by the number of the possible permutations of the output-set, that is, divide by n!.

The number of choices has to be an integer, which completes the proof.
The easiest way to see it is to use the "combinatorial" definition of $\binom mn$.

The binomial coefficient $\binom mn$ may be defined in two ways. One way is algebraic:
$$\binom mn=\frac{m!}{n!(m-n)!}.$$
This lets you manipulate binomial coefficients, and is quite useful (for example) in solving the Putnam problem. Another way, which is usually how $\binom mn$ is defined (as far as I can recall), is as "the number of ways to choose $n$ objects from a set of $m$ distinct objects, with the order of the objects chosen being irrelevant." For example, $\binom 32=3$, as there are three ways to choose two objects from a set of $3$: if the set is $\{A,B,C\}$, then you could choose $A$ and $B$, or $A$ and $C$, or $B$ and $C$. Using this definition, it's obvious that $\binom mn$ should be an integer -- it counts the number of ways to do something.

The utility of binomial coefficients is that these definitions are *equivalent*. In other words, one can show that the number of ways to choose $n$ objects from a set of $m$ distinct objects, without caring about the order in which the objects are chosen, is exactly
$$\frac{m!}{n!(m-n)!}.$$
This allows one to use either the combinatorial or algebraic definition at any point, and leads to a very clean solution of the Putnam problem (among many other applications). A proof is as follows.

1. Consider first the problem of choosing the $n$ objects and caring about the order. Then, there are $m$ ways to choose the first object, $m-1$ ways to choose the second (since you can choose any object of the $m$ besides the object which was chosen as the first one), all the way through $m-n+1$ ways to choose the last object. So, the number of ways to do this is
$$m(m-1)(m-2)\cdots(m-n+1)=\frac{m!}{(m-n)!}.$$
2. Now, let $A$ be the number of ways to do this choosing without caring about order. The number of ways to do this choosing with caring about order is $A\cdot n!$: first choose the elements, then order them in any way. This tells us that
$$A\cdot n!=\frac{m!}{(m-n)!},$$
and so $A=\frac{m!}{n!(m-n)!}$, as desired.