Let's choose 9 out of 12 chips

It can be generalised...

Suppose that we have a total of $NK$ chips, with $N$ chips of each of $K$ different colours. The number $A_M$ of ways of choosing $M$ chips is the coefficient of $x^M$ in the polynomial $F(X) \; = \; (1 + X + X^2 + \cdots + X^N)^K$ (the combination of $a_1$ of colour $1$ , $a_2$ of colour $2$ , and so in corresponds to selecting $X^{a_1}$ from the first bracket, $X^{a_2}$ from the second bracket, and so on). We note that $F(X) \; = \; \left(\frac{1 - X^{N+1}}{1-X}\right)^K \; = \; \left(\sum_{a=0}^K (-1)^a\binom{K}{a}X^{a(N+1)}\right)\left(\sum_{b=0}^\infty \binom{b+K-1}{K-1}X^b\right)$ It is particularly easy to determine $A_M$ provided that $M \le N$ , for then we only need the $a=0$ term from the first bracket, and $A_M \; = \; \binom{M+K-1}{K-1}$ The formula for $M > N$ gets more complicated. For example $A_{N+1} \; = \; \binom{K+N}{K-1} - K \hspace{2cm} A_{2N} \; = \; \binom{2N+K-1}{K-1} - K\binom{N+K-2}{K-1}$ For this particular problem we have $M=N=3$ and $K=4$ , so that $A_3 = \binom{6}{3} = 20$

It is interesting to note that $F(X)$ is a polynomial of degree $KN$ , but we have expressed it as an infinite power series. This means that we have automatically proved a variety of combinatorial identities from the fact that $A_{M} = 0$ for all $M > NK$ . For example $\sum_{a=0}^K (-1)^a \binom{K}{a} \binom{(K-a)(N+1)+K-1}{K-1} \; = \; A_{K(N+1)} \; = \; 0$