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What is a binomial coefficient?
How do binomial coefficients relate to combinatorics?
What is the significance of the binomial coefficient?
Let's do this by induction on $m$, where $n$ is fixed. We see right away that for $m = 0$, we have $$ (-1)^0 \binom{n}{0} = 1 = (-1)^0 \binom{n-1}{0}. $$
Now assume the identity holds for all values $\leq m$, and we will show that it for $m + 1$. As a heads up, we will make use of the binomial recursive formula, sometimes referred to as Pascal's Triangle: $\binom{n}{k} = \binom{n-1}{k-1} + \binom{n - 1}{k}$.
\begin{align*} \sum\limits_{k = 0}^{m+1}(-1)^k \binom{n}{k} &= (-1)^{m+1}\binom{n}{m+1} + \sum\limits_{k = 0}^{m}(-1)^k \binom{n}{k} \\ &= (-1)^m \binom{n-1}{m} + (-1)^{m+1} \binom{n}{m+1} \text{ by the inductive hypthosis}\\ &= (-1)^{m+1} \left( \binom{n}{m+1} - \binom{n-1}{m} \right) \\ &= (-1)^{m+1} \binom{n-1}{m+1} \text{ by Pascal's Triangle}. \end{align*} This completes the proof.
EDIT: A fair amount of identities with Binomial Coefficients can be proven using induction and Pascal's Triangle; if you come across a similar identity, induction and Pascal's Triangle are often a good place to start.
Here is a direct proof as well. By Binomial theorem we have $$(1-x)^{n-1}=\sum_{k=0}^{n-1} \binom{n-1}{k}(-1)^k x^k.$$ At the same time, using geometric progression, we have
$$(1-x)^{n-1}=\frac{1}{1-x}(1-x)^n=(1+x+x^2+\dots)\sum_{k=0}^{n} \binom{n}{k}(-1)^k x^k.$$
Comparing coefficients at $x^m$ on boths sides, we get
$$
\binom{n-1}{m}(-1)^m=1\cdot \binom{n}{m}(-1)^m+1\cdot \binom{n}{m-1}(-1)^{m-1}+\dots+1\cdot \binom{n}{0}(-1)^{0}
$$
You can do this by elementary combinatorics, think about how products of this kind are expanded. You need to find all the possible (ordered) products of $x$, $y$ and $z^{-1}$ with 4 factors that give $xyz^{-2}$. In order to obtain $xyz^{-2}$ one of the factors needs to $x$, one needs to be $y$ and two need to be $z^{-1}$, so we are looking for all the permutations of $(x,y,z^{-1},z^{-1})$. In general, there are $4!=24$ permutations of a $4$-tuple, but since two of the entries are equal, we are left with $4!/2=12$ different permutations. Now you just multiply the coefficients of these monomials with how often the desired combination appears in the expansion and obtain $$12\cdot x\cdot(-2)y\cdot3z^{-1}\cdot3z^{-1} = -216xyz^{-2}.$$
This is the reasoning that leads to multinomial expansion in the end, but I find it a lot easier to remember how this works, instead of shooting up the multinomial formula if I only need a single coefficient.
$$\begin{align*} \left(x-2y+3z^{-1}\right)^4 =& \left[(x-2y)+3z^{-1}\right]^4\\ =& \sum_{i=0}^4 \binom4i (x-2y)^{4-i}\left(3z^{-1}\right)^i\\ =& \binom42 (x-2y)^2\left(3z^{-1}\right)^2 + \cdots\\ =& \binom42 \left(x^2-4xy+4y^2\right)\left(3z^{-1}\right)^2 + \cdots\\ =& \binom42 \left(-4xy\right)\left(3z^{-1}\right)^2 + \cdots\\ \end{align*}$$ Other non-$xyz^{-2}$ terms are omitted.
I'm working through Oscar Levin's introduction to discrete mathematics as a self-learner without much of a background in mathematics and I'm confused by some of the exercises in the section on Binomial Coefficients.
Specifically question 8:
What is the coefficient of x12 in (x+2)15
(answer: (15 choose 12) multiplied by 23)
I understand where 15 choose 12 comes from, as there are 15 possible places in the expansion where x could be multiplied with another x term and it needs to do so in 12 of these. I can see (or at least thought I could see) how this relates to other similar problems covered in the chapter like finding the number of bitstrings with length n and weight k etc. What I don't understand, and what doesn't seem to be covered explicitly in any of the preliminary material is why having found 15 choose 12, you'd then need to multiply this by 23.
What's more, this seems to conflict with the answer to question 9:
What is the coefficient of x9 in the expansion of (x+1)14+x3(x+2)15
(answer: (14 choose 9) + (15 choose 6) multiplied by 29)
Here the coefficient of x9 in (x + 1)14 is given as simply (14 choose 9), not multiplied by anything, despite this looking like exactly the same kind of problem as in question 8. Then, for the coefficient of x9 in x3(x+2)15, we're back to multiplying by 29.
In all these cases I can follow where the values for n and k in n choose k are coming from. I can also see that multiplying by 2n-k is common to all these problems, and I can vaguely see where that might come from, insofar as n-k is the number of locations in which x isn't multiplied by x and in each location x either is or isn't multiplied by x But I'm struggling to see how this all ties together, or to find a pattern explaining why you sometimes need to multiply n choose k by 2n-k and why sometimes just n choose k seems to be enough.
Thanks in advance, any help much appreciated.