Why does the Binomial Theorem use combinations and not permutations for its coefficients?

Start with a red ball (R), a green ball (G), and an orange ball (O). N = 3

The combinations of just 2 balls (n = 2) would be RG RO GO and this works with 3!/[2!(3 - 2)!] = 3.

Flipping a coin results in either H or T.

Think of the red ball as the first flip of the coin, the green ball as the second flip of the coin, and the orange ball the third flip of the coin. But now we also add which two will come up heads. The combinations are RG RO GO.

Actually, you are right that Binomial is a permutation but with repeated letter.

For example, how many ways are there to arrange the letters $abc$? $3!$.

For $aba$, how many ways to arrange them? $3!/(2!cdot 1!)$. We need to divide due to the two $a$s, which are indistinguishable from each other.

Hence, this answer is the same formula as the combination formula, $n!/ ((n-r)!cdot r!)$.

The use of combinations in the coefficients DOES convey order, but it is indirect. Thus, it would make sense for combinations to be present. Here's why:

Consider: $(a+b)^3 = (a+b)(a+b)(a+b)$ Note: There are 3 factors, this is important

Now, how many $a^2b$ terms are there? 3

This 3 can be thought of as the result of a repeated permutation of 2a's and one b, aab, aba, aab

This 3 is also a result from combinations. i.e. From the 3 factors of $(a+b)$, which 2 am I going to choose for the a. If you chose the first 2 factors, then the term will be aab, if you chose the first and third factors, the term would be aba. However choosing the first and third factor, is the same as choosing the third and first factor and that's why it is a combination.

Conclusion: The combination involved is to do with the factors of the entire binomial $(a+b)$, not the individual a and b's! Just switch your perspective slightly :)

My way of seeing the binomial formula is the following. Suppose you want to compute $$ (a+b)^n $$ for some $ngeq 1$. Look at it in this way: $$ underbrace{(a+b)cdot (a+b) cdot ldots cdot (a+b)}_{n text{terms}}, $$ with exactly $n$ multiplications. How do you obtain the result? Pick one term between $a$ or $b$ from each factor and multiply them together. So the result contains terms of the form $a^kb^{n-k}$ for $k=0,,ldots,n$: this means that you picked $k$ times $a$ and $n-k$ times $b$. How many choices do you have? You have $n$ "different" $a$'s and you have to count the number of ways to select $k$ of them. The order does not matter: this means that if you selected the same $a$'s in a different order, the would give exactly the same term in the result, thus you don't want to count them twice. This is why for each $k$ you have exactly $binom{n}{k}$ choices.

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 $n$,

$$(a+b)^n = sum_{k=0}^n binom{n}{k}a^k b^{n-k}$$

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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.