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How many permutations of the letters of the word "DEFEATED" are there in which the E's are separated from each other?

I relied on the fact that: (The number of permutations of the letters in this word with NO restriction) - (The number of permutations of the letters in this word with the 3 E's are sticking together) = The required number of permutations.

The number of permutations of the letters in this word with NO restriction = 8! / (2!3!) = 3360

The number of permutations of the letters in this word with the 3 E's are sticking together = 6! / 2! = 360 (in here I regarded the 3 E's as 1 item as a whole so (EEE)DFATD are 6 items in total with 2 identical letters (D) so the result is 6! / 2!).

Now, doing the subtraction gives 3000 which should be the answer to the question but this is not correct according to the answer given in the book. What might be wrong in my method?!

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    $\begingroup$ Perhaps E's are not considered to be separated, if two E's are sticked together and the third one is separated from these two... $\endgroup$ – 042 Jul 6 '12 at 12:16
  • $\begingroup$ You have to subtract the case when two E's are together and 1 E is separated also. $\endgroup$ – Aang Jul 6 '12 at 12:17
  • $\begingroup$ Inclusion Exclusion principle might help $\endgroup$ – sabertooth Jul 6 '12 at 12:17
  • $\begingroup$ See math.stackexchange.com/q/162764, where exactly the same error was committed for a rather similar problem. $\endgroup$ – Marc van Leeuwen Jul 6 '12 at 13:22
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My preferred method for this goes as follows:

First work out the number of permutations of the letters which are not E: this is just $5!/2$ (didvide by 2 for the number of ways of rearranging the identical Ds).

Now imagine the these letters written something like this: .D.F.A.T.D.

If the Es are to be separated, then we have to choose 3 of the dots to be the places where the Es can go.

If you work out the number of ways of choosing 3 from 6, and then multiply by $5!/2$, you should get the right answer.

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  • $\begingroup$ Since 2 of the 5 not-E letters are identical, you need to divide that 5! by 2, no? $\endgroup$ – Gerry Myerson Jul 6 '12 at 13:00
  • $\begingroup$ Ah yes - I will correct it $\endgroup$ – Old John Jul 6 '12 at 13:01

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