Which would be correct: $$P_4=4!=24 \quad \text{or} \quad \frac{4!}{1!2!1!}=12?$$

I don't know if Meеt and Meеt with exchanging the two $e$'s are different or not? If the answer is the latter, why would it be different from the permutations of the word Meat with an $a$, which should be $24$ permutations I believe?


  • 1
    $\begingroup$ Wow, I don't know if this is how I should use the comment section but I got to say everyone clarified this for me perfectly.. I tried to accept 2 answers at the same time at one point lol. Thanks all! $\endgroup$
    – Viper
    Dec 11 '17 at 8:19

This is not a matter of mathematics. You already gave the correct answer to your own question: It depends if you treat the two $e$'s as identical (then you have $12$ permutations) or as distinguishable (in this case you have $24$ permutations). Mathematics can't help you to decide which one fits for you.

If the word is written on a computer, using any standard text editor, then both $e$'s are identical, even if you look deep into the electronic storage. This gives you $12$ permutations.

But if you hand out colored pens to a child, ask him or her to write the word on a piece of paper, and then cut out the four letters and shift them around on your table, you will easily be able to distinguish the two versions of the letter $e$. So now you will find $24$ permutations.

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    $\begingroup$ +1, though you might as well also mention that the intended answer is certainly 12. $\endgroup$
    – ruakh
    Dec 12 '17 at 4:42

You can think like this:

If the letters would be different then you would have $4!$ as the number of permutations, but in the word $MEET$ you have two letters that are indistinguishable so swapping these two doesn't give rise to a new permutation. So we have to "remove" these, and the number of ways $2$ letters can be permutated is $2!$ so we have to divide with $2!$.

Hope this helps :)


The word "meat" will have $^4P_4=24$ permutations. Consider two of these $24$ permutations, i.e
$p_1 = $ "mtae" and $p_2 = $"mtea", if we replace 'a' with 'e' in both $p_1$ and $p_2$ then they will become identical i.e "mtee". Thus two distinct permutation of the word "meat" corresponds to same permutation upon replacing 'a' with 'e'. Hence we need to divide $24$ by $2$ to get the right result. If $n$ letters are identical then the internal permutation of these $n$ letters corresponds to identical permutation of the whole word and we must then divide by $n!$.


Let's ask Python and itertools.permutation!

If 'e's are identical, you get a collection of $\frac{4!}{1!2!1!}$ unique strings:

>>> from itertools import permutations
>>> sorted(set(''.join(chars) for chars in permutations('meet')))
['eemt', 'eetm', 'emet', 'emte', 'etem', 'etme', 'meet', 'mete', 'mtee', 'teem', 'teme', 'tmee']
>>> len(set(''.join(chars) for chars in permutations('meet')))

If 'e's are distinguishable, e.g. with a stroke on one 'e', the set has $4!$ elements:

>>> sorted(set(''.join(chars) for chars in permutations('meɇt')))
['emtɇ', 'emɇt', 'etmɇ', 'etɇm', 'eɇmt', 'eɇtm', 'metɇ', 'meɇt', 'mteɇ', 'mtɇe', 'mɇet', 'mɇte', 'temɇ', 'teɇm', 'tmeɇ', 'tmɇe', 'tɇem', 'tɇme', 'ɇemt', 'ɇetm', 'ɇmet', 'ɇmte', 'ɇtem', 'ɇtme']
>>> len(set(''.join(chars) for chars in permutations('meɇt')))

Instead of generating every permutation and filtering the duplicates, it's also possible to generate the list directly without any duplicate. Here's an example.

It generates $\frac{10!}{9!}$ ($=10$) strings without having to generate $10!$ strings and removing $10*(9!-1)$ duplicates:

>>> [''.join(chars) for chars in perm_unique('aaaaaaaaab')]
['baaaaaaaaa', 'abaaaaaaaa', 'aabaaaaaaa', 'aaabaaaaaa', 'aaaabaaaaa', 'aaaaabaaaa', 'aaaaaabaaa', 'aaaaaaabaa', 'aaaaaaaaba', 'aaaaaaaaab']
>>> len(list(permutations('aaaaaaaaab')))

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