Proof of binomial coefficient formula. How can we prove that the number of  ways choosing $k$ elements among $n$ is    $\frac{n!}{k!(n-k)!} = \binom{n}{k}$    with $k\leq n$?
This is an accepted fact in every book but i couldn't find a proof.
Notice that given $n$ objects $\{a_i\}$ and $n$ positions, there are $n$ possible positions for $a_n$. Once we have placed $a_n$, there are $n-1$ remaining positions for $a_{n-1}$. Once we have placed $a_n$ and $a_{n-1}$, there are $n-2$ positions left for $a_{n-2}$, and so forth. Thus, there are
$$
n(n-1)(n-2)\cdots3\cdot2\cdot1=n!
$$
possible ways to arrange the $n$ items.

Suppose we divide the $n$ items into $k$ white objects and $n-k$ black objects. For each arrangement of the white and black objects that looks the same, there are $k!$ rearrangements of the white objects and $(n-k)!$ rearrangements of the black objects. Since there are $n!$ rearrangements of the $n$ objects, there must be
$$
\frac{n!}{k!(n-k)!}=\binom{n}{k}
$$
arrangements of the $k$ white and $n-k$ black objects that look different.
If you have 5 people (A, B, C, D &amp; E) but wants to build a team of three people, the most number of ways you can do is 5x4x3= 60 ways. But realize that team ABC and CBA is basically same team.
So for three people, you can have 3x2x1=6 different combinations.
Meaning that, making a team with 5 poeple with 3 positions, you can have total (5x4x3) / (1x2x3) = 10 different combinations. This number 10 is the 3rd number on 5th row of Pascal's triangle. This basically tells us the number of combination we can have.
And (5x4x3) / (1x2x3) = 5! / (3! x 2!) = C (5, 3)