Obtaining binomial coefficients without "counting subsets" argument

Question

I want to obtain the formula for binomial coefficients in the following way: elementary ring theory shows that $(X+1)^n\in\mathbb Z[X]$ is a degree $n$ polynomial, for all $n\geq0$, so we can write

$$(X+1)^n=\sum_{k=0}^na_{n,k}X^k\,,\ \style{font-family:inherit;}{\text{with}}\ \ a_{n,k}\in\mathbb Z\,.$$

Using the fact that $(X+1)^n=(X+1)^{n-1}(X+1)$ for $n\geq1$ and the definition of product of polynomials, we obtain the following recurrence relation for all $n\geq1$:

$$a_{n,0}=a_{n,n}=1;\ a_{n,k}=a_{n-1,k}+a_{n-1,k-1}\,,\ \style{font-family:inherit;}{\text{for}}\ k=1,\dots,n-1\,.$$

I want to know if there is a way to manipulate this recurrence in order to obtain directly the values of the coefficients $a_{n,k}$, namely $a_{n,k}=\binom nk=\frac{n!}{k!(n-k)!}$.

Note that the usual approach via generating functions definitely will not work, at least in the spirit of my question, because this method only works when we do know in advance the coefficients of the generating function (either by the "number of $k$-subsets" argument, or Maclaurin series for $(X+1)^n$, or anything else) and this is precisely what I intend to avoid.

This question is closely related to a recent question of mine. Actually the same question, with Bernoulli numbers instead of binomial coefficients.

EDIT

I do not consider as a valid manipulation the following "magical" argument: 'the sequence $(b_{n,k})$ given by $b_{n,k}=\frac{n!}{k!(n-k)!}$ obeys the same recurrence and initial conditions as $(a_{n,k})$, so $a_{n,k}=b_{n,k}$ for all $n,k$. How did you obtain the formula for the $b_{n,k}$ at the first place? Okay, you can go through the "counting subsets" argument, but this is precisely what I don't want to do. The same applies to my related question about Bernoulli numbers.

Answer 1

A simple-minded approach is to solve the two variable recurrence relation iteratively, that is, knowing $a_{n,0}$ find $a_{n,1}$, then $a_{n,2}$, etc.

We must have
$$\begin{eqnarray*} a_{n,1} &=& a_{n-1,1}+a_{n-1,0} \\ &=& a_{n-1,1}+1, \qquad a_{1,1}=1. \end{eqnarray*}$$ This is a one variable recurrence relation of the form $$b_n = b_{n-1}+1, \qquad b_1 = 1.$$ This can be solved by the usual methods. We find $a_{n,1} = n.$

Next we have $$\begin{eqnarray*} a_{n,2} &=& a_{n-1,2}+a_{n-1,1} \\ &=& a_{n-1,2}+n-1, \qquad a_{2,2}=1. \end{eqnarray*}$$ This is another simple recurrence relation. We find $a_{n,2} = n(n-1)/2.$

At the next step, $$\begin{eqnarray*} a_{n,3} &=& a_{n-1,3}+a_{n-1,2} \\ &=& a_{n-1,3} + \frac{1}{2}(n-1)(n-2), \qquad a_{3,3}=1. \end{eqnarray*}$$ This implies $a_{n,3} = n(n-1)(n-2)/6.$

This process can be repeated to build up $a_{n,k}$ for any $k$. At some point a pattern will be noticed and the principle of induction can be applied.

Answer 2

If you define $a_{n,k}=\binom nk$ to be the coefficient of $X^k$ in $(1+X)^n$ (with no combinatorial meaning attached), then you easily find $\binom nk=\binom {n-1}{k-1}+\binom{n-1}k$ for all $n,k\geq1$ (as you indicated in the question). Also $\binom n0=1$ for all $n\geq0$, and $\binom0k=0$ for all $k>0$. Expanding the second term of the recurrence relation recursively, with $\binom0k=0$ as terminating clause, gives $$ \binom n k=\sum_{0\leq i<n}\binom i{k-1} \qquad\text{for $k\geq1$ and all $n\geq0$.} $$ Now from the fact that $n\mapsto\binom n0$ is a polynomial function (of $n\in\mathbf N$) of degree$~0$ (namely the constant function$~1$) it follows easily by induction that $n\mapsto\binom nk$ is a polynomial function of degree$~k$ (summing a polynomial function of$~n$ increases the degree by$~1$). Moreover the first $k$ values $\binom ik$ for $0\leq i<k$ of the polynomial function are all$~0$, so the polynomial has roots $0,1,2,\ldots,k-1$. It therefore necessarily takes the form $\binom nk=c_k(n-0)(n-1)\ldots(n-(k-1))$ for some constant $c_k$. Using the recursion (at some point one does need to compute something) the first nonzero term $\binom kk$ of the sequence $n\mapsto\binom nk$ can be shown to be$~1$ for every$~k$, and it follows that $c_k=1/(k(k-1)(k-2)\ldots(k-(k-1))=\frac1{k!}$. Then $$ \binom nk = \frac{n(n-1)\ldots(n-(k-1))}{k!}, $$ which you may, if you feel so inclined, write as $\binom nk = \frac{n!}{k!(n-k)!}$ as long as $k\leq n$.

Look mom, no counting!