Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Given n letters 'A' and n letters 'B', how many unique words can be obtained using all 2n letters? Or how many unique arrangements exist for those letters? Why is the answer $C_{2n} ^n = \frac{2n!}{n!n!}$? I understood the fact that given 2n letters there are 2n! permutations and that because it's the same letter repeated there are identical permutations, but how do you reach that formula?

And in general, I've been told by my teacher than given a number of letters, the general formula would be $\frac{n!}{n_1! ... n_k!}$ where n is the total number of letters and $n_k$ the number of letters of the K-th letter. Why is that?

share|cite|improve this question
Hint: Such a word is determined entirely by the position in which the $n$ As occur. – Thomas Andrews Feb 7 '13 at 18:27
up vote 3 down vote accepted

Let's assume for a moment that all $A'$s and $B'$s are distinct, say $A_1,A_2,\ldots A_n$ and same goes for the $B's$. One of the $2n!$ word formed by those letters would be $$ A_1A_2\cdots A_n B_1B_2\cdots B_n. \,\, (1) $$ At first, this word would be different than the word $$ A_nA_{n-1}\cdots A_1B_1B_2\cdots B_n, $$ but since in fact all the $A'$s are alike, all the words of the form $(1)$ are counted $n!$ times (fixing the $B'$s and permutting the $A's$. ) Since this is also true for the $B'$s, we get another $n!$ factor to divide our $2n!$ with.

share|cite|improve this answer
And I assume the same kind of reasoning goes for the generalized case with k letters. – andreas.vitikan Feb 8 '13 at 4:48
Yes exactly, just replace each of my $n!$ by the numbers $n_1!,n_2!,\ldots n_k!$ where each $n_i$ is the number of the letter $"i"$. – Jean-Sébastien Feb 8 '13 at 16:06

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.