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In how many ways can the letters of the word SOCKS be arranged in a line so that the two S's are together? In how many arrangements can the letters in SLOOPS be arranged so that the two O's are together?

I would think the answer to the first one would be: Treat the two S's as one entity and permute the letters: 4! and divide by 2! to account for the identical element S.Apparently not. However for the second question, you are able to use this method?

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    $\begingroup$ For the first there is no need to divide by $2!$ because the two S's are combined entity. In the second case you need to divide by $2$ because of the two S's. Think if two S's were different, then $S_1LOOPS_2$ and $S_2LOOPS_1$ would be considered different. But they are same so you need to compensate for that overcounting by dividing by $2$. $\endgroup$ – Anurag A Dec 21 '16 at 19:34
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For the first question, you are arranging $\fbox{SS}\text{OCK}$, where $\fbox{SS}$ is a single entity, so there are $4$ different elements and the permutations work out to $4!$ without further adjustment.

For the second question, you are arranging $\fbox{OO}\text{SLPS}$, where again $\fbox{OO}$ is a single entity, so there are $5$ elements of $4$ different values, one of which is repeated, and the permutations work out to $5!/2! = 60$, the division by $2!$ adjusting for the repeated S.

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  • $\begingroup$ So are you saying if you were to have the two O's to be separated, you would say there are 10 different positions where the O's can be apart and the remaining 4 letters permuted but as to account for the S's you would have: 10*4!/2! ? $\endgroup$ – TripleA Dec 21 '16 at 20:06
  • $\begingroup$ If the Os were separable you would have $6$ letters ($6!$ options) but with two sets of 2-fold repeats (divide by $2!$, twice) for $6!/(2!2!) = 720/4 = 180$ options (if you don't care whether the Os are together or apart). Which gives the same answer (via $180-60$) of $120$ for requiring the Os to be separate. $\endgroup$ – Joffan Dec 21 '16 at 22:17

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