# Stuff I forgot 45 years ago - Permutations

ok, so daughter is taking a college Statistics class.

questions was Permutation of the Letters in COMMITTEE

So it is 9 letters, with 3 repeats of 2 = 9! / (2! 2! 2!) = 362,880 / 8 = 45360

What confuses me is ignoring the single letters. I guess I could try to derive it, but anyone have a simple explanation?

You could include the single letters in the denominator, but they would each end up being 1!, which is just 1.

but if all individual N! / (N-R)!

but in this formula, denominator instead of N-R, it seems to add up the Ns
ok. got it, no real R. If 9 individual letters (with repeats) it is just 9!

thanks, needed to talk it out

I would agree with this answer for distinguishable permutations (but it has been over 55 years since I last determined permutations and combinations).

You might start with the idea that each letter is a different color; so you have 9 unique letter-color items to select for a permutation: For example, you swap an orange M with the green M and you have a different order of the elements.

So there are 9! ways to chose those 9 letters.

Now, if all of the letters are the same colorâ€“equivalently, weâ€™re going to ignore color nowâ€“changing the position of the two Mâ€™s is no longer a different combination. So how many â€śduplicatesâ€ť would there be? Well, consider the case when the letters were different colors and youâ€™ll see (after some thought) that youâ€™ll have 2! x 2! x 2! possible ways of selecting the order of the â€śduplicatesâ€ť for any given configuration where those letters would appear in a sequence. So we just divide these out from the total.

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