0

I don't know if this has been asked before.

Consider the word $ABBD$. Now, suppose we want to find the number of $2$-permutations of the letters. As per my knowledge, the number would be $$\frac {4!}{2!.2!}=\frac {4.3}{2!}$$ Is this correct? If no, then please provide with a correct expression, if no, here's my question

Are not the words $AD$ and $DA$ undercounted?

For, any 2 permutation including the repeated letters are divided by $2$, but in doing this, we also include the words that do not have the repeated letters. Where am I wrong?

Eisenstein
  • 435

1 Answers1

1

The formula $\frac{4!}{2!2!}$ does not apply here. It would apply for choosing two letters from $ABCD$, where order does not matter. But you want to choose from $ABBD$, where order does matter, so this is unrelated. To answer one of your questions, yes, $AD$ and $DA$ are undercounted, since you are ignoring order when you divide by $2!$ a second time.

There is no general formula for "partial anagrams" of a word. See this question and the related questions for the closest thing to a formula. However, since this problem is small, you can just count all possibilities directly.

Mike Earnest
  • 84,902