Say I have 3 A's and 2 B's, which make up a set of 5 numbers.
How many permutations are there of that set?
Doing it by hand, I can see that there are 10, but with my relatively limited math knowledge, I haven't really been able to see any obvious pattern when you change the numbers.
What's the formula that would tell me how many permutations there are?
Preface
This is a simple question, and uses simple binomial calculation. However, it can be generalized. Here, I shall provide two possible solutions. Depending on the situation of the problem, as well as its complexity, you might choose to use which method.
Solution 1
The number of ways to choose 3 out of 5 slots for the A's is $\binom {5}{3} =10$ ways
That is the answer to this problem.
The solution can be generalized as following: Given a set of k character, consider the couple $(a_i,b_i)$ with $a_i$ refers to the character and $b_i$ represents the number of appearances.
Then, the answer is as follow:
$ \binom{b_1+b_2+...b_k}{b_1} + \binom{b_2+b_3+...+b_k}{b_2} + ... + \binom{b_k}{b_k}$
Solution 2
I first assume that the five characters given are different, then I have $5!=120$ ways to arrange them.
Now because the A's are the same, and the B's are the same, changing their situations does not matter. Therefore, the number of repetitions are: $3! \times 2! =12$
This means that actually, there are only: $ \frac{120}{12}=10$ ways to arrange the numbers.
The generalization is as following:
Denote the couples $(a_i,b_i)$ as in solution 1
Denote $S= \sum_{i=1}^{k} b_i$
The total number of solutions is:
$ \frac{S!}{b_1! \times b_2! \times b_3! \times... \times b_k!}$
Post script: I left here two possibles solutions to this question, as it depends on the situation that you chooses to use which solution to the generalised problem