IMMEDIATE

We are looking for arrangements of the form T _ _ _ _ _ _ _ A . If the letters in between were all distinct, we would have $7! = 5040$ possibilities. However, the letters $I, M, \text{and } E$ occur twice. For each of these, the possibilities are counted twice (for example the arrangement T _ E _ E _ _ _ A ) occurs twice in 5040). Therefore, we need to divide the total possibilities by 2 for each of these letters and the answer is $\frac{7!}{2^3} = \frac{5040}{8} = 630$ .

For $T$ in first and $A$ in last, we have to give up them from arrangement. Because they are state and they have a fixed position.

Then we have $2I$ , $2E$ , $2M$ and $1D$ for arrange

So we get the total arranges are $=\frac{7!}{2!\times2!\times2!}=\boxed{630}$