1. Count the letters in EXAMINATION

The word EXAMINATION has 11 letters:

$E, X, A, M, I, N, A, T, I, O, N$

So the repeated letters are:

• $A$ occurs $2$ times
• $I$ occurs $2$ times
• $N$ occurs $2$ times

All others ($E, X, M, T, O$) occur once.

Thus we have:

• $5$ letters repeated twice: actually only the types repeated are $A, I, N$ i.e. $3$ types
• $5$ distinct single-occurring letters: $E, X, M, T, O$

So total distinct letter types $= 8$.

2. We need 4-letter words

Since no letter appears more than twice, the possible repetition patterns in a 4-letter word are:

1. All 4 letters distinct: $(1,1,1,1)$
2. One letter repeated twice, two other distinct letters: $(2,1,1)$
3. Two letters repeated twice each: $(2,2)$

We count each case separately.

3. Case 1: All 4 letters distinct

Choose 4 distinct letters from the 8 distinct letter types, then arrange them.

$\binom{8}{4}\cdot 4! = 70 \cdot 24 = 1680$

4. Case 2: Pattern $(2,1,1)$

Choose the letter that is repeated twice. This repeated letter must be one of $A, I, N$.

So number of choices for repeated letter:

$3$

Now choose 2 other distinct letters from the remaining 7 distinct letter types:

$\binom{7}{2} = 21$

Now arrange the multiset of letters of type $(2,1,1)$:

$\frac{4!}{2!} = 12$

Hence total for this case:

$3 \cdot \binom{7}{2} \cdot \frac{4!}{2!} = 3 \cdot 21 \cdot 12 = 756$

5. Case 3: Pattern $(2,2)$

Choose 2 letters from the 3 repeated letter types $A, I, N$:

$\binom{3}{2} = 3$

Arrange the letters of type $(2,2)$:

$\frac{4!}{2!2!} = 6$

Hence total for this case:

$\binom{3}{2} \cdot \frac{4!}{2!2!} = 3 \cdot 6 = 18$

6. Total number of 4-letter words

$1680 + 756 + 18 = 2454$

7. Comparison with stored correct answer

Our derived answer is:

$\boxed{2454}$

This matches the stored correct answer.