$\begin{array}{1 1}(A)\;907200\\(B)\;917200\\(C)\;926200\\(D)\;93630\end{array} $

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- $n!=n(n-1)(n-2).....(3)(2)(1)$

Words starting with A are formed with the letters 2I's,2N's A,E,X,M,T,O

Numbers of words formed by these letters $=\large\frac{10!}{2!2!}$

$\Rightarrow \large\frac{10\times 9\times 8\times 7\times 6\times 5\times 4\times 3\times 2\times 1}{2\times 2}$

$\Rightarrow 10\times 9\times 8\times 7\times 6\times 5\times 3\times 2\times 1$

$\Rightarrow 907200$

Then the words starting with $E,I,M,N,O,T,X$ will be formed.

$\therefore$ Number of words before the first word starting with E is formed

$\Rightarrow 907200$

Hence (A) is the correct answer.

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