Permutation on word if E,F,G have to stay in order Im stuck on a problem which I have answered and need help to verifiy if I have done/understood it correctly.
Problem
If we have the following string: A,A,B,C,D,D,D,E,F,G
How many ways are there to rearrange the letters if the Letters, E,F,G have to appear in the same order.
My approach
I first decided to find the permutation of the letters except E,F,G and deal with these 3 last. So i ended up with $\frac{10\cdot9\cdot8\cdot7\cdot6\cdot5\cdot4}{2!\cdot3!}$ 
And now that there are 3 more letters left. (The last three) and that they have to appear in order, we have no choice other than placing the letters on order. i.e E,F,G
So the final answer is $\frac{10\cdot9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot1\cdot1\cdot1}{2!\cdot3!}$
Question
I would like to know if I have attempted this problem correctly. If not could you point out where i have gone wrong and/or give me some hints?
Edit
E,F,G doesn't have to appear in next to eachother
 A: The number of ways to arrange all $10$ letters is found from the multinomial coefficient as: 
$$\binom{10}{3,2,1,1,1,1,1} = \frac{10!}{3!2!1!1!1!1!1!}=302,400$$
Then for each ordering of all $10$ letters if we think of keeping all but those in the set $S=\{E,F,G\}$ fixed then there are $3!$ ways to arrange the elements in $S$.  If we divide the above number by $3!=6$ we get the answer you're looking for which is $50,400$.
A: You said the E, F, and G have to appear in order. Do they have to appear consecutively?
If so, then your approach started out correct; finding all the possible 7-letter words of the remaining letters and inserting the string "EFG" in various places would yield the answer. However, you would want 7-letter words:
$\cfrac{7\cdot6\cdot5\cdot4\cdot3\cdot2}{3!\cdot2!} = 420$
(so you would not start the multiplication with $10$)
After this, there are eight places to insert the string "EFG" (before the first letter, between the first and second letters, between the second and third, etc.)
So $8\cdot420 = 3360$
If E, F, and G do NOT have to appear consecutively, the problem becomes a bit more difficult.
A: I assume that order of E,F,G would be same and E,F,G can appear anywhere in the 10 letter string, not necessarily consecutively.
So, First select 3 out of 10 positions, where to put E,F,G that would be
$$\binom{10}{3} = \frac{10.9.8}{3.2} = 120$$
Further, permuting 7 alphabets gives you 420.
So the answer would be $120*420 = 50400$
A: First permute the non-ordered letters in the multiset $\{A^2, B^1, C^1, D^3\}$, which gives you
$$\binom{7}{2, 1, 1, 3} = \frac{7!}{2! 1! 1! 3!} = 420$$
Next there are 8 positions for the other 3 letters, you need to pick 3 of them, for
$$\binom{8}{3} = \frac{8!}{3! 5!} = 56$$
I.e., in all:
$$\binom{7}{2, 1, 1, 3} \binom{8}{3} = 420 \cdot 56 = 23520$$
