$\begin{array}{1 1}(A)\;360\\(B)\;192\\(C)\;96\\(D)\;48\end{array} $

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

Total number of words starting with C=5!=120

Which include the word COCHIN also.

$\therefore$ The number of words that appear before the word COCHIN is less than 120.

Now total no. of words starting with CC=$4!=24$

Similarly in each of the cases total word starting with CH,CI,CN is 4!=24

$\therefore$ Total no. of words before starting with 10=$4\times 24=9!$

The first word with CO in alphabetical order is COCHIN.

Hence (C) is the correct answer.

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