Solution:

Start with an empty string that should have 5 letters. We first choose where to put $I$ and $E$, and there are $\binom{5}{2}$ ways of doing this. Each way has $3!$ ways to rearrange the remaining letters, so the number of ways is $10 \times 6 = 60$ ways.

Alternatively, note that the letters can be arranged without restriction in $5!$ ways. Since there are the same number of arrangements where $I$ appears before $E$ as those where $E$ appears before $I$, the required number is $5! / 2 = 60$ ways.