What I did was to separate out the 3 E's (EEE) from the word ELEMENTARY, which leaves L M N T A R Y. For no two E's to be next to each other, we can place the E's in 8 positions (between the alphabets
L M N T A R Y and also before L and after Y). Firstly, since all the letters L M N T A R Y are distinct the number of ways to permute them is 7! = 5040. Then to place the 3E's in the 8 locations; firstly the first E have 8 choices to choose from, then second E has 7 choices to choose from and 3rd E has 6 choices to choose from. Then using product rule, 5040 x 8 x 7 x 6 = 1693440.
However, my answer is wrong, the method above is correct till the "7! = 5040". In the solution that was given to me, there are indeed 8 positions to place the 3 E's. However, instead of using 8 x 7 x 6, the solution used C(8,3) = 56. Then, using product rule, 5040 x 56 = 282240 ways of permutating ELEMENTARY such that no two E's are next to each other.
Why is my answer wrong? Thank you! :")
Best Answer
As you state, the difference between your answer is that you multiply by $8*6*5$ whereas the correct answer multiplies by $C(8,3)$. By definition, $C(8, 3) = \frac{8!}{3! 5!} = \frac{8*6*5}{3!}$, so your approach missed thus division by $3!$.
Essentially, your approach counted the same configuration multiple times. Let's label these 3 E's as $E_1, E_2, E_3$. Then your argument correctly shows that there are $7! * 8 * 7 * 6$ ways to arrange the letters of $E_1LE_2ME_3NTARY$ so that no $E_i$'s are adjacent. But of course, we aren't working with $E_i$, we're just working with $E$! Your counting would consider $E_1LE_2ME_3NTARY$ and $E_2LE_1ME_3NTARY$ to be different but all we did was shuffle the same letter around. That is, permuting the 3 E's doesn't lead to a new configuration. And how many permutations of 3 elements are there? $3!$ (And I'm not just excited!)
To summarize, for every correct configuration of $ELEMENTARY$, your method counts each $E_{\sigma(1)}LE_{\sigma(2)}ME_{\sigma(3)}NTARY$ as distinct while they should be the same, for $\sigma$ a permutation of $\{1,2,3\}$. There are $3!$ such permutations so you have to divide your result by $3!$ to get the correct one.