Hi,
I understand where you got your original answer, but as pointed out, it is wrong. Here's one way to count. List all the possibilities for first and last letter and for each count the number of arrangements of the remaining 7 letters.
1. a..a: 7!/4 2 m's and 2 t's in the remaining letters
2. a..e: 7!/4 2 m's and 2 t's in the remaining letters
3. a..i: 7!/4 2 m's and 2 t's in the remaining letters
4. e..a: 7!/4 2 m's and 2 t's in the remaining letters
5. e..i: 7!/8 -- there are 2 a's, 2 m's and 2 t's in the remaining letters
6. i..e: 7!/8 -- there are 2 a's, 2 m's and 2 t's in the remaining letters
7. i..a: 7!/4 2 m's and 2 t's in the remaining letters
The sum of the arrangements 1 through 7 is then
$${6\cdot 7!\over 4}$$
There are 9 letters with 2 A's, 2 T's, 2 M's
Vowels are A, A, E, I
The start and end can have AA, 2xAE, 2xAI, 2xEI
With AA 7!/{2! 2!] = 1260
With 2xAE = 2 x 7!/[2! 2!] = 2520
With 2xAI = 2 x 7!/[2! 2!] = 2520
With 2xEI = 2 x 7!/[2! 2! 2!] = 1260
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Total = 7560 <-----
Many Thanks. I also got the answer.
Here is how I would do that: If we were to treat the two "A"s, two "M"s, and two "T"s as different (write one as " " and the other " " and similarly for the "M"s and "T"s) then there are 4 choices for the leading vowel. That leaves 3 choices for the final vowel. That leaves 7 letters that can be any where in the middle so 7! ways that can be done. That is there are (4)(3)(7!) ways to order the letters in " ".
But interchanging the "A"s does NOT give a new order. Treating the two "A"s as the same we have to halve that. Similarly for the two "M"s and two "T"s. The correct number is .