Binomial Coefficients in the Binomial Theorem - Why Does It Work Question to keep it simple: Given
$(a+b)^3=\binom{3}{0}a^3+\binom{3}{1}a^2b+\binom{3}{2}ab^2+\binom{3}{3}b^3$
Could you please give me an intuitive combinatoric reason to why the binomial coefficients are here?
for instance, what does $\binom{3}{2}ab^2$ mean combinatorially?
Thank you very much.
 A: $\binom{3}{2}$ means the number of possibilities to choose two elements from a three-element set without replacement. More generally, $\binom{n}{k}$ means how many ways there are to chose a $k$-element subset of an $n$-element set.
Now, $(a+b)^3 = (a+b)(a+b)(a+b)$, so there are three factors; let's call them $1,2,3$. Now, when expanding this product into a sum, we must consider all possible pairings of summands. Since they all have the same form, we can basically just use a subset of $\{1,2,3\}$ to denote from which factors we take the $a$; the rest of the factors will then be $b$.
So, taking a subset $S \subseteq \{ 1,2,3 \}$, we get a term that takes the $a$ terms from the factors listed in $S$, and the $b$ terms from the factors not in $S$. Some further thought shows that this just gives a term $a^{|S|} b^{3-|S|}$. Now, there are exactly $\binom{3}{|S|}$ subsets of size $|S|$, so you get $\binom{3}{|S|}$ terms of the form $a^{|S|}b^{3-|S|}$. The binominal theorem follows from this, since you need to consider all subsets.
A: $\binom{3}{2}ab^2=abb+bab+bba$
A: This isn't as rigorous as the accepted answer, but explores the question a bit from another perspective.
Take your example,
$(a+b)^3=\binom{3}{0}a^3+\binom{3}{1}a^2b+\binom{3}{2}ab^2+\binom{3}{3}b^3$
Let us consider this small rewrite of the same example:
$(a+b)^3=\binom{3}{0}{a}\cdot{a}\cdot{a}+
\binom{3}{1}{a}\cdot{a}\cdot{b}+
\binom{3}{2}{a}\cdot{b}\cdot{b}+
\binom{3}{3}{b}\cdot{b}\cdot{b}$
Consider that the binomial if expanded would take on the following product:
$(a+b)^3 = (a+b)(a+b)(a+b)$
To compute this product we have to pick one summand in each of the $(a+b)(a+b)(a+b)$ and multiply them together.  We then do this in all possible ways adding each term together.  Since there's $2$ possibilities for each of $3$ choices there's $2^3=8$ total possibilities.  In essence, we are forming all length 3 strings over the alphabet $\{a,b\}$.
That may be hard to follow, but it's easy to see it written out:
$(a+b)^3 = (a+b)(a+b)(a+b) = $
${a}\cdot{a}\cdot{a} + 
{a}\cdot{a}\cdot{b} + {a}\cdot{b}\cdot{a} + {b}\cdot{a}\cdot{a} +
{b}\cdot{b}\cdot{a} + {b}\cdot{a}\cdot{b} + {a}\cdot{b}\cdot{b} + {b}\cdot{b}\cdot{b}$
So there's $8$ possibilities, so let's think about it intuitively as you asked:


*

*How many length 3 strings can we make over $\{a,b\}$ where we are allowed to choose $0 b$'s?  Well, there's $\binom{3}{0}$ of those: $\{aaa\}$.  Hence the first term, $\binom{3}{0}a^3$.

*How many length 3 strings can we make over $\{a,b\}$ where we are allowed to chose $1 b$?  Well, there's $\binom{3}{1}$ of those: $\{aab,aba,baa\}$.  Hence the second term, $\binom{3}{1}a^2b$.

*Same as above, but we choose $2 b$'s? $\{bba,bab,abb\}$. To get third term, $\binom{3}{2}ab^2$.  This is what Marc van Leeuwen's terse answer is conveying, and is in essence the "combinatorial meaning" you asked for in your example.

*Now we choose $3 b$'s? $\{bbb\}$. To get fourth term, $\binom{3}{0}b^3$.

Dave L. Renfro makes a good point:

none of the current answers explain why (a+b)(a+b)(a+b) can be
  expanded by considering the sum of all possible ordered products of
  three elements, where the first element comes from the first factor of
  (a+b), the second element comes from the second factor of (a+b), and
  the third element comes from the third factor of (a+b).

I'll do my best to explain this one.  You see, we're actually repeatedly applying the distributive law of multiplication over addition.
For this example, consider this:
$(a+b)(a+b)(a+b)$
$= ((a+b)a + (a+b)b)(a+b)$  Here you distribute the first factor into the second.
$= (aa + ba + ab + bb)(a+b)$  Distribute again.
$= (aa(a+b) + ba(a+b) + ab(a+b) + bb(a+b))$ Distribute again.
$= aaa + aab + baa + bab + aba + abb + bba + bbb$
Intuitively as we distribute an $(a+b)$ we are "choosing" a or b, but recording both results... in a way.
In effect if we say distribute $c$ over $(a+b)$ we are effectively recording what happens to c when we choose a, and also when we choose b, e.g. $c(a+b) = ca + cb$
