Consider the pascal triangle of powers of $\,(a+b)$ :
\begin{array} {lc} n&(a+b)^n\\ \hline 0&1\\ 1&a+b\\ 2&a^2 + 2ab + b^2\\ 3&a^3 +3a^2b+3ab^2+b^3\\ 4&a^4 +4a^3b+6a^2b^2+4ab^3+b^4\\ \end{array} To go from $1$ to (say) $a^2b$ using only multiplication by $a$ or $b$ you may use different paths (from top to bottom) :
- $1\to a\to a^2\to a^2b\ $ (we choose $b$ at the third bifurcation)
- $1\to a\to ab\to a^2b\ $ (we choose $b$ at the second bifurcation)
- $1\to b\to ab\to a^2b\ $ (we choose $b$ at the first bifurcation)
From the three possible paths we will add three terms and get $3$ in front of $\,a^2b$.
If you forget $a$ (i.e. set $a:=1$) you may think that you choose or not to multiply by $b$ at each of the three iterations corresponding to one choice between three : $\,\large{\binom{3}{1}}$.
To get to $\,ab^2\,$ instead you would have to choose $b$ two times between three : $\,\large{\binom{3}{2}}$ and so on.
Except for $a^n$ we may come from the upper left or the upper right only so that the total number of paths from $1$ to $\,a^{n-k}b^k\;$ (in $(a+b)^n$ with $b$ to power $k>0$) will be given by :
$\{$number of paths to $a^{n-k}b^{k-1}\}+\{$number of paths to $a^{n-k-1}b^k\}$ : $\,\displaystyle\binom{n}{k}=\binom{n-1}{k-1}+\binom{n-1}{k}$
that is $\large{\binom{3}{1}=\binom{2}{0}+\binom{2}{1}}\,$ in our $a^2b$ example.