To solve this problem, we can count the total number of words possible and subtract the number of words where the two letters 'E' are next to each other.
Let's start by finding the total number of words possible if there are no restrictions. The word 'EXAMPLE' has 7 letters, so there are 7 positions to fill. For the first position, we have 7 choices (any letter can go there). For the second position, we have 6 choices (any letter except the one used in the first position). Similarly, for the third position, we have 5 choices, and so on. Therefore, the total number of words without any restrictions is:
7 choices for the first position × 6 choices for the second position × 5 choices for the third position × 4 choices for the fourth position × 3 choices for the fifth position × 2 choices for the sixth position × 1 choice for the seventh position = 7! (7 factorial)
Now, let's count the number of words where the two letters 'E' are next to each other. We can treat the two 'E's as a single entity. So, we have six entities: E, X, A, M, P, L. The number of words with the two 'E's together can be found in the same way as before, but now we have only 6 entities instead of 7. Therefore, the number of words where the two 'E's are together is:
6 choices for the first position × 5 choices for the second position × 4 choices for the third position × 3 choices for the fourth position × 2 choices for the fifth position × 1 choice for the sixth position = 6!
Finally, we subtract the number of words where the two 'E's are together from the total number of words without any restrictions to obtain the number of words where the two 'E's are not next to each other:
7! - 6! = 5040 - 720 = 4320
Therefore, there are 4320 different words possible where the two letters 'E' are not next to each other when using the letters from 'EXAMPLE'.