Divide the entire problem into all possibilities. Let's consider one by one.

Length of the word is 5. So, the general format will be $n*\frac{5!}{(repeating \:elements)!} $

$where \:n \:is\:how \:we\:can\:choose\:elements.$

1. $\textbf{All 5 are same: }$ Only a,b,c,d and e are 5 or more than 5. Hence we choose 1 out of 5. So,

$\binom{5}{1} * \frac{5!}{5!} $

2. $\textbf{4 are same and 1 is different:}$ 7 elements are more than or equal to 4. The remaining 1 can be any out of the 8 (as we choose 1 already which will occur 4 times). So,

$\binom{7}{1} \binom{8}{1} * \frac{5!}{4!*1!}$

3. $\textbf{3 are same and other 2 are same:}$ All 9 alphabets are more than or equal to 3. Hence, we can choose any 1 out of 3. Now, even if we choose here alphabet who is of least frequency ($\textit{say i}$), we'll be left with at least 2 alphabets. Hence, it is safe to say the 2 elements can be chosen out of the remaining 8. So,

$\binom{9}{1} \binom{8}{1} * \frac{5!}{3!*2!}$

4. $\textbf{3 are same and rest 2 are different:}$ Choose 1 out of 9 and rest 2 from 8

$\binom{9}{1} \binom{8}{2} * \frac{5!}{3!*1!*1!}$

5. $\textbf{2 are same, 2 are same(but different from previous), and 1:}$ Choose any 1 out of 9 which will occur only once. The rest 2 which will occur twice can be chosen from the remaining 8.

$\binom{9}{1} \binom{8}{2} * \frac{5!}{2!*2!*1!}$

6. $\textbf{2 are same, rest 3 are all different:}$ Choose one that occurs twice out of 9. Rest 3 out of remaining 8.

$\binom{9}{1} \binom{8}{3} * \frac{5!}{2!*1!*1!*1!}$

7. $\textbf{All 5 are different:}$ Simply choose 5 out of 9. Order will matter and hence we will take permutation.

$^9P_5$

If you add all these, you'll get 58965

If you are wondering how to get the cases, they are simply partitions of 5.

[5] [4,1] [3,2] [3,1,1] [2,2,1] [2,1,1,1] [1,1,1,1,1]