How many $3$-Letter Strings from come from $ABRACADABRA$? How many three-letter strings are there that come from the word $ABRACADABRA$?
I know I can split this in to cases, but I was hoping that there would be a better way of solving it.
 A: Here is an approach using exponential generating functions.  Whether this is an improvement over case by case analysis depends on how you feel about algebra and maybe how you feel about using a computer algebra system.  Personally, I use a CAS and feel no guilt.
More generally, let's see if we can find how many $r$-letter words can be formed; call this number $a_r$.  Define the exponential generating function of $a_r$ by
$$f(x) = \sum_{r=0}^{\infty} \frac{1}{r!} a_r x^r$$
Then it's "easy to see" (ahem) that
$$f(x)=(1+x+\frac{1}{2!}x^2 +\frac{1}{3!}x^3 +\frac{1}{4!}x^4 +\frac{1}{5!}x^5) \cdot (1+x+\frac{1}{2!}x^2)^2 \cdot (1+x)^2$$
If we expand this (probably using a CAS), we will have the solution to the problem for all values of $r = 0, 1, 2, \dots , 11$.  But for your problem, all we really need is the coefficient of $\frac{1}{3!} x^3$.  It's possible to find this coefficient by pencil and paper without a lot of work, but doing so is very similar to case by case analysis, so I'm not sure that is an improvement over the case by case approach.
Using a CAS, it turns out that the coefficient of $x^3$ is 97/6, so the coefficient of $\frac{1}{3!} x^3$ is 97.  That's your answer.
A: My approach:
Case 1. Strings of the form $AAA$:
1
Case 2: Strings of the form $AAB$:
$\binom{3}{1} * 4 * \frac{3!}{2!} = 36$
Case 3: Strings of the form $ABC$:
$\binom{5}{3}*3! = 5*4*3 = 60$
Total = $1+36+60=97$
A: I don't see a direct approach, but I could divide the problem in $3$ cases:
(1) When all the letters are distinct, then there are $5\cdot 4\cdot 3$ strings (there are exactly $5$ distinct letters).
(2) When exactly two (of the three) letters are identical, they they can be only $3$ ways of choosing these letters (A, B or R), then the remaining letter can be any of the $4$ remaining letters which then can be positioned in any of the $3$ positions of the string (so, we obtain $3\cdot 4\cdot 3$ strings of this type)
(3) When all $3$ letters are identical, then they all must be A's (so, there is only $1$ string of this type).
Adding the results of the $3$ cases gives the answer $97$.
A: I think that it is like that:
$$\binom{11}{3} \cdot \frac{11!}{5! \cdot 2! \cdot 2!}$$
You want a string of $3$ letters,so from the 11 possible letters you choose three ones.
But you have to consider that there are $5$ $A$s, $2$ $B$s and $2$ $R$s.
