The number of 4 letter words (with or without meaning) that can be formed from the eleven letters of the word EXAMINATION is

First, let us analyse the word EXAMINATION. It contains 11 letters in all, but several of them repeat. Listing each distinct letter with its frequency, we have

$$E_1,\; X_1,\; A_2,\; M_1,\; I_2,\; N_2,\; T_1,\; O_1$$

Thus the eight different letters and their available copies are

$$E(1),\; X(1),\; A(2),\; M(1),\; I(2),\; N(2),\; T(1),\; O(1).$$

We wish to form all possible 4-letter “words” (ordered arrangements) without exceeding these individual limits. Order matters, so the task is to count all 4-permutations drawn from this multiset.

Because no letter occurs more than twice in the original word, the possible repetition patterns inside any 4-letter word are only

$$\;(1,1,1,1),\quad (2,1,1),\quad (2,2).$$

Patterns like $$(3,1)$$ or $$(4)$$ are impossible, since no letter appears three or four times in the source.

We now treat each admissible pattern separately and add the results.

Case I : all four letters different $$(1,1,1,1)$$

We first choose 4 distinct letters out of the 8 available. Using the combination formula $$\binom{n}{r}= \dfrac{n!}{r!\,(n-r)!},$$ we get

$$\binom{8}{4}=70.$$ Once the 4 distinct letters are chosen, they can be arranged in $$4! = 24$$ different ways. Hence

$$\text{words in this case}=70 \times 24 = 1680.$$

Case II : one letter repeated twice and two other different letters $$(2,1,1)$$

(a) Choosing the letter that appears twice: only the letters A, I and N possess two copies, so we have $$3\ \text{choices}.$$

(b) After fixing that repeated letter, 7 other distinct letters remain (because 1 of the 8 has already been used). We must pick any 2 of these, so

$$\binom{7}{2}=21.$$ (c) For a multiset containing two identical and two different letters, the permutation count is obtained by the division rule: $$\frac{4!}{2!}=12.$$

Therefore

$$\text{words in this case}=3 \times 21 \times 12 = 756.$$

Case III : two letters, each appearing twice $$(2,2)$$

(a) Both repeated letters must again come from the set {A, I, N}. Selecting any 2 of them gives

$$\binom{3}{2}=3.$$ (b) A multiset of the form $$a,a,b,b$$ has $$\frac{4!}{2!\,2!}=6$$ distinct permutations.

Hence

$$\text{words in this case}=3 \times 6 = 18.$$

Adding all cases together

$$\begin{aligned} \text{Total words} &= 1680 + 756 + 18 \\ &= 2454. \end{aligned}$$

So, the answer is $$2454$$.