So I have this word (a word doesn't have to mean anything):
AAABBCCDDDEEE
Finding the number of anagrams is easy (correct me if I am wrong): $$\frac{(3+2+2+3+3)!}{3!2!2!3!3!}$$
Now the question is: how many words with the same number of anagrams are there?
This was a question present on an exam and I don't have any idea on how to approach it. Thank you for your time!
On
Hint: Assuming that words must be formed out of the 26 letters in the alphabet, focus on how many distinct letters are in this word, and how many ways there are to swap one letter with another.
On
The answer comes in two parts
The easier part is the number of "words" with the same pattern of three letters appearing three times each and two letters appearing two times each, and so (if your alphabet has $26$ letters) twenty-one letters appearing zero times each. This is $\frac{26!}{3!2!21!}$, perhaps multiplied by the answer you already have
The harder part is considering whether there are other patterns which might generate the same answer you already have. It might help to note your original word has $13$ letters
A really interesting question!
To find the number of possible patterns if we know the total number of letters is a nice problem which can be solved with a reasonably straightforward procedure.
General notation
The general formula for the number of anagrams of an $N$ letter word is $$\frac{N!}{1!^a2!^b3!^c...}$$
Example If the word has 13 letters
Then we require $$2!^b3!^c4!^d= 2!^23!^3$$ [Note that we cannot have any factorial greater than $4!$ since $5$ does not divide the RHS.]
Equating the total number of letters $$a+2b+3c+4d=13$$ Equating the powers of 2 $$b+c+3d=5$$ Equating the powers of 3 $$c+d=3$$
There are just two solutions. $(a,b,c,d)$ is either $(0,2,3,0)$ (the original pattern) or $(3,0,2,1)$.
How many letters can a possible word have?
As noted by @Henry the word must have at least $13$ letters. Since the prime $17$ does not divide $13!$, the word can only have $17$ or more letters if it had $17$ copies of the same letter ... there is no solution of this type but my proof is long-winded!
If the word has 14 letters
$$\frac{14}{1!^a2!^b3!^c4!^d5!^e6!^f7!^g...}=\frac{1}{2!^23!^3}$$
$7!$ must occur on the denominator to cancel out the factor of 7 in the 14. But then 5 also divides the denominator and this is not possible.
If the word has 15 letters
$$\frac{15.14}{1!^a2!^b3!^c4!^d5!^e6!^f7!^g...}=\frac{1}{2!^23!^3}$$
The only solution is $$(a,b,c,d,e,f,g)=(2,0,2,0,0,0,1)$$
If the word has 16 letters
$$\frac{16.15.14}{1!^a2!^b3!^c4!^d5!^e6!^f7!^g...}=\frac{1}{2!^23!^3}$$
The only solution is $$(a,b,c,d,e,f,g)=(1,0,0,2,0,0,1)$$
So, barring any errors, there are 4 different patterns.