# AAABBBCCDE Combinations

1. Jan 15, 2009

### nobahar

1. The problem statement, all variables and given/known data
Hello!
Okay, using the letters AAABBBCCDE, draw three letters. What is the probability of getting exactly two letters the same?

2. Relevant equations AND 3. The attempt at a solution
I worked my way through the possible combination (e.g. AAB, AAC, AAD, etc.) Then used combinations: $$\left(\begin{array}{c}3\\2\end{array}\right)$$ x $$\left(\begin{array}{c}3\\1\end{array}\right)$$ + $$\left(\begin{array}{c}3\\2\end{array}\right)$$ x $$\left(\begin{array}{c}2\\1\end{array}\right)$$ + $$\left(\begin{array}{c}3\\2\end{array}\right)$$ x $$\left(\begin{array}{c}1\\1\end{array}\right)$$, etc.

To calculate the probabilities, I get the right answer, but I was wondering if there is an easier way of identifying how many possible combinations there are. For example, if I had to find how many different combinations there are when each letter is different; then, ABCDE gives me 5C3, 10 combinations to work through, I still have to write them out to see what I'm working with, but I know how many I have, whereas I don't necessarily know when the selection of three involves exactly two of the same letters, this is even more important when the range of letters, symbols, objects, etc are many more in number!
Any help appreciated. Thanks

EDIT: I thinks it's just: 3 x (4 x 1 x 1), since I have one way of choosing the first, one for the second, four for the third, and this applies to all the letters that can possible appear twice. Seems a bit tedious, especially if there is a large selections of possibilities...

Last edited: Jan 15, 2009
2. Jan 15, 2009

### Defennder

Re: Combinations

Hint: What is the probability of not choosing the same letter twice? So that leaves us with the alternatives of choosing either all 3 same letters or only 1 same latter, the latter simply equates to choosing all three distinct alphabets.

3. Jan 18, 2009

### nobahar

Re: Combinations

Thanks!
I assuming that looking at the complement events is the best way then.