Zoom International Math League

The two "E"s make things a little tricky here. Note that in a rearrangement the "E"s are identical, as swapping them doesn't change anything.

Let's look at a smaller example, were we can list everything. How many different rearrangements of "ROAR" are there that contain "OAR" in the sequence.

Here there's only $2! = 2$ (not $2!\cdot 2 = 4$) different rearrangements that contain "OAR":

1. ROAR
2. OARR

I think it also helps to list out ALL the different rearrangements of "ROAR" in total (note there are not $4! = 24$ rearrangements, there are $4!\div 2 = 12$ of them:

1. AORR
2. AROR
3. ARRO
4. OARR
5. ORAR
6. ORRA
7. RAOR
8. RARO
9. ROAR
10. RORA
11. RRAO
12. RROA

Hope this helps!