Using Exponential Generating Functions on Counting Problems Is it possible to use exponential generating functions to solve problems where repetition is wanted?
For example, if I wanted to solve the following problem which wants distinct possibilities...

How many different 5-letter words can be formed from the letters from the word ABRACADABRA if duplicated letters are allowed but no letter can be used more times than it occurs in the word ABRACADABRA ? 
5 distinct letters, A repeated five times, B twice, C once, D once, R twice.
Then, the function would be
  $$ F(x) = (1 + \frac {x}{1!} + \frac {x^2}{2!} + \frac {x^3}{3!} + \frac {x^4}{4!} + \frac {x^5}{5!}) (1 + \frac {x}{1!} + \frac {x^2}{2!})^2 (1 + \frac {x}{1!})^2 $$
  $$ F(x) = \frac {x^{11}}{120} + \frac {3 x^{10}}{40} + \frac {2 x^9}{5} + \frac {19 x^8}{12}+ \frac{289 x^7}{60}+\frac{331 x^6}{30}+\frac{2221 x^5}{120}+\frac {545 x^4}{24} + \frac{121 x^3}{6}+\frac{25 x^2}{2}+5 x+1 $$
Therefore, the answer would be $5!(\frac {1271}{120})$

However, how would I be able to apply it to the following problem?

How many 5-card hands (from an ordinary deck) have at least one card of each suit? 
4 suits, repeated 13 times
$ 52*51*50*49*48 = 311,875,200 $ total possibilities
$$ G(x) = (\frac {x^{13}}{13!} + \frac{x^{12}}{12!}+ \frac{x^{11}}{11!} + \frac{x^{10}}{10!} + \frac{x^9}{9!} + \frac{x^8}{8!} + \frac{x^7}{7!} + \frac{x^6}{6!} + \frac{x^5}{5!} + \frac{x^4}{4!} + \frac{x^3}{3!} + \frac{x^2}{2!} + \frac{x}{1!} + 1)^4 $$ 

Of course, this would just give me unique combinations, so it doesn't work. Is there an actual method to solve using exponential generating functions? Is it possible or even recommended?

The generating function for the second problem will provide you with all possible 5-card hands. But we are interested in the case when we have at least one cord of each suit. So we need to make some restrictions. We already know the suit of 4 of the cards (because the conditon states they are different) and there are 4 options for the fifth card, so there are 4 different combinations if the cards were undistiguishable. But because they aren't, we need to multilply that number by the multinomial:
$$\binom{5}{2,1,1,1} = \frac{5!}{2!\cdot 1! \cdot 1! \cdot 1!} = 60$$
So the total number of combinations is:
$$60 \cdot 4 = 240 \text{ combinations}$$
This is actually the number of possible combinations with a pair of cards of the same suit. We can get the same exact result using the exponential generating function. Actually the exponential generating function is based on the same principle. You can read more about this here, especially on the $7^{th}$ page.
But before applying the exponential generating function, we need to limit the number of cards of one suit. From the condition we can obtain that the maximum number of cards per suit is 2 and the minimum number of cards per suit is 1. So the GF will be:
$$G(x) = \left(x + \frac {x^2}{2!}\right)^4 = x^4+2 x^5+\frac{3 x^6}{2}+\frac{x^7}{2}+\frac{x^8}{16}$$
The coefficient in fron to $x^5$ is 2. So the total number of combinations is:
$$2 \cdot 5! = 2 \cdot 120 = 240 \text{ combinations}$$
And we can see that we've obtained the same results using both methods. That's because, the exponetnial generating function is a more elegant, hidden and simplier version of the multinomial method. 
