# Anagrams with $n$ letters and Generating Functions.

Find an exponential generating function for $${a_r}$$ , the number of $$r$$- letter words with no vowel used more than once (consonants can be repeated).

The answer is $$(1 + x)^5e^{21x}$$.

One solution I see is the following:

Since there can be one $$(x^1 = x)$$ OR no vowel $$(x^0 = 1)$$, the generating function for the vowels, by the additive principle, is $$1 + x$$. The exponential generating function for the consonants is $$e^x =\sum_{n\ge 0}\frac{x^n}{n!}$$. Since there are $$21$$ consonants, we have $$e^{21x}$$. By multiplying the two, we have: $$(1 + x)^5e^{21x}$$. My doubts are:

1) Why the generalized function for each of the consonants is $$e^x$$?

2) Why do we multiply it $$21$$ times?

3) Why, in the end, do we multiply both?

The number of $$r$$-letter words which have only constants is $$21^r$$. Hence the exponential g.f. is $$\sum_{r=0}^{\infty}\frac{21^r x^r}{r!}=e^{21x}.$$ Now consider the case of the $$r$$-letter words with a vowel "a". Their number is $$r21^{r-1}$$ (why?) and the exponential g.f. is $$\sum_{r=1}^{\infty}\frac{r21^{r-1} x^r}{r!}=x\sum_{r=1}^{\infty}\frac{21^{r-1} x^{r-1}}{(r-1)!}=xe^{21x}.$$ So the e.g.f. of the $$r$$-words with consonants and at most one "a" is $$\sum_{r=0}^{\infty}\frac{(21^r+r21^{r-1})x^r}{r!}=(1+x)e^{21x}.$$ Can you take it from here?