If the different permutations of all the lett | vedclass

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(B) The word $EXAMINATION$ consists of $11$ letters: $A, A, I, I, N, N, E, X, M, T, O$.In a dictionary,words are arranged alphabetically. The letters

Vedclass · Accepted Answer

$907200$

Vedclass · Answer

(B) The word $EXAMINATION$ consists of $11$ letters: $A, A, I, I, N, N, E, X, M, T, O$.In a dictionary,words are arranged alphabetically. The letters in the word are $A, E, I, M, N, O, T, X$.The words starting with $A$ will appear before the words starting with $E$.To find the number of words starting with $A$,we fix $A$ at the first position.The remaining $10$ letters are $A, I, I, N, N, E, X, M, T, O$.Among these $10$ letters,$I$ appears $2$ times and $N$ appears $2$ times.The number of permutations of these $10$ letters is $\frac{10!}{2!2!} = \frac{3628800}{4} = 907200$.Thus,there are $907200$ words starting with $A$ before the first word starting with $E$.

If the different permutations of all the letters of the word $EXAMINATION$ are listed as in a dictionary,how many words are there in this list before the first word starting with $E$?

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