how many words are there of 5 letters that have at least one I and at least two T's, but no K or Y?

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So it is more of a riddle than research-level mathematics, but your help would be hugely appreciated.

Here is the full problem: With the Latin alphabet, how many words are there (counting all of them, even those that don't make sense) of 5 letters that have at least one I and at least two T's, but no K or Y?

I already know the answer because I coded it, but I'm looking for simple math proof. Here is my code.

import itertools

alphabet = string.ascii_uppercase.replace("K", "").replace("Y", "")

count = 0
for word in itertools.product(alphabet, repeat = 5):
    if "I" in word and word.count("T") >= 2:
        count += 1

print (count)

The code answer is 15645

Thank you so much!

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prerequisite :

the number of permutation with n objects of which a are alike , b are alike , c are alike ... is

$$\frac{n!}{(a!*b!*c!*....)}$$

proof:

3 letters for the 5 letter word are fixed i.e (i,t,t)

lets signify all letters other than i,t,k,y using x

for the 2 letters left , we can have seven cases:

case 1 : i,i

case 2 : i,t

case 3 : i,x

case 4 : t,t

case 5 : t,x

case 6 : x,x but both letters are different

case 7 : x,x but both letters are the same

as these are all the cases that exist , the sum of all these cases will give us the answer

case 1 :

the set contan i,i,i,t,t

no of possible permutatons : $$\frac{5!}{(2!*3!)} = 10$$

case 2 :

the set contain i,i,t,t,t

no of possible permutatons :

$$\frac{5!}{(2!*3!)} = 10$$

case 3 :

the set contain i,i,t,t,x (x represent any of the 22 letters)

no of possible permutations : $$\frac{(22*5!)}{(2!*2!)} = 660$$

case 4 :

the set contains i,t,t,t,t

no of possible permutations : $$\frac{5!}{4!} = 5 $$

case 5 :

the set contains i,t,t,t,x

no of possible permutations : $$\frac{(22*5!)}{3!} = 440$$

case 6 :

the set contains i,t,t,x,x (with both x reprsent different letter )

no of possible permutations : $$\frac{({22 \choose 2} *5!)}{(2!)} = 13,860$$

case 7 :

the set contains i,t,t,x,x (both x representing same letter )

no of possible permutations : $$\frac{(22*5!)}{(2!*2!)} = 660$$

sum of all cases : 10 + 10 + 660 + 5 + 440 +13,860 + 660 = 15,645