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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You have 11 alphabets in the word PESSIMISTIC:
- three S
- three I
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
xpositions. 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.