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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Solution

In this problem, we have to find the number of words starting with A. In EXAMINATION, we have two I's and two N's and all other letters are different.

$∴$ Number of ways of arrangement = $\frac{10!}{2! 2!}$

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

= $907200.$

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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?

In this problem, we have to find the number of words starting with A. In EXAMINATION, we have two I's and two N's and all other letters are different. ∴ Number of ways of arrangement = 10!2! 2! = 10×9×8×7×6×5×4×3×2!2×1×2! = 907200.