The reason combinations come in can be seen in using a special example. The same logic applies in the general case but it becomes murkier through the abstraction.
Consider
$$(a+b)^3$$
If we were to multiply this out, and not group terms according to multiplication rules (for example, let $a^3$ remain as $aaa$ for the sake of our exercise), we see
$$(a+b)^3 = aaa + aab + aba + baa + abb + bab + bba + bbb$$
Notice that we can characterize the sum this way:
$$(a+b)^3 = (\text{terms with 3 a's}) + (\text{terms with 2 a's}) + (\text{terms with 1 a}) + (\text{terms with no a's})$$
(You can also do the same for $b$, the approach is equivalent.) Well, we see from our weird expansion that we have every possible sequence of length $3$ made up of only $a$'s and $b$'s. We also know some of these terms are going to group together, as, for example, $aba = aab = baa$.
So how many summands are actually equal? Well, since they all have the same length, two summands are equal if and only if they have the same number of $a$'s (or $b$'s, same thing). And we also know that every possible sequence of length $3$ and only $a$'s and $b$'s are here.
So we can conclude that
$$\begin{align}
(\text{# of terms with 3 a's}) &= \binom{3}{3} = 1\\
(\text{# of terms with 2 a's}) &= \binom{3}{2} = 3\\
(\text{# of terms with 1 a}) &= \binom{3}{1} = 3\\
(\text{# of terms with no a's}) &= \binom{3}{0} = 1
\end{align}$$
Thus, we conclude:
- There will only be one $aaa = a^3$ term
- There will be $3$ $aba=aab=baa=a^2b$ terms.
- There will be $3$ $abb = bab = abb = ab^2$ terms.
- There will be $1$ $bbb=b^3$ term.
Thus,
$$(a+b)^3 = \sum_{k=0}^3 \binom{3}{k}a^k b^{3-k}$$
and in general, for positive integers $n$,
$$(a+b)^n = \sum_{k=0}^n \binom{n}{k}a^k b^{n-k}$$
In short, the reason we use combinations is because the order does not matter, because we will get terms like $aab, baa, bab$ which are all equal in the expansion. Since multiplication is a commutative operation over the real numbers, then, we can say they're equal. Thus, the number of terms of that "type" (characterized by how many $a$'s or $b$'s they have) is given precisely by the number of sequences of length $n$ ($n=3$ in our example), made of only $a$ and $b$, that has exactly $k$ $a$'s (or $b$'s).
Of course this all relies on the central premise that multiplication commutes in the reals and thus ensures that the order of the factors does not matter. That suggests that it does not always hold in situations where multiplication does not commute - for example, the multiplication of a type of numbers known as quaternions is not commutative, and thus the binomial theorem does not hold there as it does here (since there $ab$ need not equal $ba$).
The nature of this commutativity, or the lack of it, and the consequences of each is better divulged in a discussion on abstract algebra, and this tangent is long enough as it is.
Best Answer
Here is the combinatorical approach:
Suppose that we choose $k$ out of $n$ elements where order does matter. We would have $n$ posibilities for the first choice, $n-1$ posibilities for the next and so on. So we have $$ n\cdot (n-1) \cdot (n-2) ... (n-k+1) = \frac{n!}{(n-k)!}$$ posibilities, but this is only when order does matter. When order doesn't matter, we would have to divide by the number of ways we can arrange the $k$ objects, which of course is $k!$. We thereby get that there is $$\frac{n!}{k!(n-k)!}$$ ways to choose $k$ elements from a set of $n$ elements (when order does not matter).