Please help

Asked by Rahulsinha1993 | 5th Nov, 2009, 10:22: AM

Expert Answer:

Let's select a letter P, and PERMUTATIONS have 12 letters.

Imagine 12 different adjacent boxes for 12 different letters.

P can go in any of the 12 boxes, 12 ways.

Then E can go in any 11. and since one box is occupied by P. So on for next letter.

So the number of words/number of ways of permuting them = nx(n-1)x(n-2)...x1 = n! = 12 ! = 479001600 Words,

But since the letter T appears twice = 239500800.

To find the 100th word according to dictionary. Let's select the letter A. Then all other letters are 11 in number and among themselves will form more than 100 words.

So let's fix the next letter as well, which is E, then permutations of 10 letters.

We find that AEIMNOP*****. and the *'s indicate the remaining 5 letters to be permuted.

5 !/2 = 120/2 = 60.

Therefore, AEIMNOR*****.

We now go backwords in the remaining permutations and find that the last permutation or 120th word has 

UTTSP as last five letters.

Hence 102th word is  AEIMNORTUTSP and 101th AEIMNORTUTPS.





Answered by  | 5th Nov, 2009, 04:59: PM

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