In how many permutations of word "LOGARITHM" vowels have alphabetical order? The answer is $9 \cdot 8 \cdot 7$ but I can't process it .

  • $\begingroup$ You want to count permutations in which the vowels are in order? If you just stick all the vowels at the front in order and then scramble the rest you already have 720, more than $9\cdot 8\cdot7$ $\endgroup$ – rschwieb May 2 '14 at 10:04
  • $\begingroup$ yes you're right , this answer is absolutely incorrect. $\endgroup$ – combine May 2 '14 at 10:08
  • $\begingroup$ Bonus points if you can solve the problem using logarithms. $\endgroup$ – DanielV May 2 '14 at 10:48

Presuming you mean just that the A has to be before the I and the I before the O in the permutation, I don't think the answer is 9*8*7.

Think about it this way: for every "good" permutation, how many "bad" ones are there where the vowels are in the same positions but our of order? For instance the good permutation ALGIROTHM has a matching bad permutation ALGORITHM.

Now if you know how many permutations there are altogether, and how many bad permutations there are for every good one, you can get the answer.


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