I'm trying to answer this question. This is my attempt of solution:

  • First we distiguish the O's and R's, then we have the word: $TO_1MO_2R_1R_2O_3W$.

  • We have $8!-7!\cdot3!-6!\cdot 3!$ different permutations of this word such that the O's aren't together. In fact, if two O's are together such as $TO_1O_2MO_3R_1R_2W$, then we have $7!\cdot3!$ permutations (the double O is viewed as a single letter). If the three O's are together such as $TO_1O_2O_3MR_1R_2W$, then we have $6!\cdot 3!$ permutations (the triple O is viewed as a single letter). Thus we subtract the total permutation $8!$ by $7!\cdot3!$ and $6!\cdot 3!$.

  • divide everything by $3!\cdot2!$ to handle with the repetitions.

  • The answer is $\frac{8!-7!\cdot3!-6!\cdot 3!}{3!\cdot2!}=480$ which is different than André Nicholas' answer.

    I would like to know why I'm wrong.


  • $\begingroup$ Can you explain how did you get second step? Also why don't just count count total words and subtract those in which O's are togethor? $\endgroup$
    – Taylor Ted
    Jul 31 '15 at 3:12
  • $\begingroup$ Can you explain $8!-7!\cdot3!-6!\cdot 3!$ $\endgroup$ Jul 31 '15 at 3:15
  • $\begingroup$ If each of the letters were different, the answer would clearly be $8!$ (8 letters). But we can order the 3 O's in $3!=2$ different ways, and the 2 R's in $2!=2$ ways, giving $\frac{8!}{2!3!}=3360$ ways, ignoring the fact that the O's cannot be together. $\endgroup$ Jul 31 '15 at 3:22
  • $\begingroup$ @trueblueanil see my edit please $\endgroup$
    – user75086
    Jul 31 '15 at 3:33
  • 1
    $\begingroup$ @user75086 The permutations for two O's togethor also constitutes of having 3 O's togethor. You have not fixed any letter between two O's and single O left over $\endgroup$
    – Taylor Ted
    Jul 31 '15 at 4:06

You are double counting, as, for example, "TOOOMRRW" is included in both your double-O case and your triple-O case. This explains why your answer is smaller than Andre's.

Consider counting your solution a little differently. We need to find the number of double-O cases that are not triple-O cases. As you said, there are 7 positions to put the double-O, since we can treat it as one letter. However, we need to make sure that the third O does not touch this double-O letter. We have to case this even further: if the double-O is on an edge, then there are 5 possible positions for the third O; otherwise, it only has 4 possible positions. Given the positions of the double-O and the third O, the remaining 5 letters can be placed in any order. Further, the 3 O's can also be shuffled.

Thus, our total for the double-O case is $(5+5+4+4+4+4+4)*5!*3!$. Our final solution is $(8! - 6!*3! - (5+5+4+4+4+4+4)*5!*3!)/(3!*2!) = 1200$.

  • $\begingroup$ Thank you! It was exactly I was looking for! $\endgroup$
    – user75086
    Jul 31 '15 at 5:02

If the question was # of permutations in which all 3 O's are not together, you could easily use Total ways - ways with all O's together

This "subtraction" method also works well for "none of the O's together" if there were only 2 O's

For more than 2 O's, it becomes quite complex, and is best avoided. What you have not considered is that when 2 O's are together, it could be any two of the 3 O's, and that the "solo" O has a number of positions where it can be placed, which must be apart from the "double O's.

But why go into all that ? Use Andre's method !


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