[Math] How many permutations of the word TOMORROW can be made if the O’s can’t be together

combinatoricspermutationssolution-verification

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 AndrĂ© Nicholas' answer.

    I would like to know why I'm wrong.

Thanks

Best Answer

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 $(8! - 6!*3! - (5+5+4+4+4+4+4)*5!*3!)/(3!*2!) = 1200$.

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