All the letters of the word PESSIMISTIC are to be arranged so that no two S's occur together, no two I's occur together, and S, I do not occur together. The number of such arrangements is


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    $\begingroup$ Please explain what you have attempted and where you are stuck so that you receive responses that address the specific difficulties you are encountering. $\endgroup$ – N. F. Taussig Mar 15 '18 at 18:15
  • $\begingroup$ I understand now that we're supposed to fix the positions of S's and I's in, say, odd places. The even positions will be occupied by the rest of the letters. This way, no 2 Ss, 2Is and SIs will be together. Thank You! $\endgroup$ – Shinjini Rana May 26 '18 at 6:15

You have 11 alphabets in the word PESSIMISTIC:

  • three S
  • three I
  • {P,E,M,T,C}

Due to the constraints in the question, the S's and I's have to take the x positions, and the remaining alphabets take the o positions.


Can you continue?

1. Fix S's position: choose three out of six x positions. This gives $\binom63$ choices. Then I's positions are fixed. 2. The remaining five letters are all different, this gives $5!$ permutations since we have five position o. To conclude, we have $\binom63 \times 5! = 2400$ possible arrangements.


You have 11 places to fill: _ _ _ _ _ _ _ _ _ _ _ In order to ensure that no two S's occur together, no two I's occur together, and no S,I occur together, it must be the case that the Ss and the Is occupy the 1st,3rd,5th,7th,9th and 11th places. Number of ways of doing so is 6!/(3!3!). Note that we divide by 3!3! to account for the fact that the three Ss are identical and so are the three Is. Finally, the remaining 5 places can be filled in 5! ways. Therefore, the total no.of ways is [6!/(3!3!)]*[5!] = 2400.


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