The letters AABBBA are arranged in a row. How many different ways can they be rearranged?
I have no idea how to do this.
I think it is 20 ways? I get this from 6!/(3!*3!). Since there are 6 letters in general, we have 6! ways of arranging them. Then, we have to account for the repeats of A by dividing by 3! (since there are 3 repeats///3 As). We divide by 3! again due to the three repeats of B for the exact same reason as with A, and that's how we get 20.
If there's no limit (e.g. 'must have an A on one end' or something), you use the factorial (!). For this, since there are 6 letters to be arranged, it would be 6!, which is 6*5*4*3*2*1.
Someone check if I'm right, though, there may be more to this question than I'm seeing, I don't know; haven't done this for a long time.
Mmm, we have to account for the fact that the same letters appear, which would make certain orders the same. For example, if we numbered each of these letters, we could have:
A
1 A
2 A
3 B
1 B
2 B
3We could also have:
A
2 A
3 A
1 B
1 B
2 B
3However, these would both be A A A B B B, making them identical arrangements. We divide by the respective factorial amounts to account for these reappearing orders.