# Find coefficient of Generating function

Here is my problem and what i have thought so far: How many ways i can arrange 9 letters when i have 3 pairs-> AA|BB|CC|DEF so without using generating functions i get $$\frac{n!}{n_1n_2n_3n_4n_5n_6n_7n_8n_9}$$ and i get$$\frac{9!}{8}$$ so how can i solve this using generating functions?

My first thought was that i use all 9 so my coefficient need to be $$coefficient*x^9$$, but i need to use egf or and ordinary and why? using an ordinary function i get $$(1+x+x^2)^3*(1+x)^3$$ , somewhere i heard about identities of genfunctions can someone enlighten me?

The exponential generating function for the number of strings of length $$r$$ taken from the given symbols is $$f(x) = \left( 1 + x + \frac{1}{2!} x^2 \right)^3 (1+x)^3$$ It's easy to see that the coefficient of $$x^9$$ in $$f(x)$$ is $$1/2^3$$, so the coefficient of $$(1/9!) x^9$$, which is the number of strings of length $$9$$, is $$\frac{9!}{2^3}$$

To make the answer more complete, I just indicate a derivation of the exponential generating function $$f(x)$$ that awkward used.

Let $$S$$ be the alphabet of symbols and we have as restriction for the number of $$s$$ symbols used, $$n_s$$: $$0 \le n_s \le k_s, s\in S$$. Then:

$$f(x)=\sum_{k=0}^{\infty} C_k \frac{x^k}{k!} =\prod_{s\in S} \sum_{n_s=0}^{k_s}\frac{x^{n_s}}{n_s!} [1]$$

One can check that the exponential function agrees with the one used by awkward, with $$S=\{A,B,C,D,E,F\}$$ and $$k_A=k_B=k_B=2$$ and $$k_D=k_E=k_F=1$$.

The expression is easily verified thinking that the number of configurations of length k and fixed occupations $$\{ n_s \}$$ is $$\frac{k!}{n_1!...n_s!}$$. Summing over all possible choices:

$$C_k=\sum_{\{n_s\}|n_1+...+n_s=k, 0\le n_s \le k_s} \frac{k!}{n_1!...n_s!},$$

which is exactly what the expansion of the right member of [1] provides.