Calculating the number of unique arrangements of the word REVERSE, taking into account the repeated Es and Rs, is straightforward:


I am not sure, however, to calculate the restriction when the letters are separated.

One source suggests that we arrange the 5 remaining letters. This creates $6$ spaces into which the remaining V and S can be permuted into:


Is this logic correct?

  • $\begingroup$ Yes, it is fine. I would prefer to count the "bad" ones (not separated) and subtract from $420$. Really, would rather use $7!-2\cdot 6!$, and at the end divide by $2!3!$. $\endgroup$ – André Nicolas Nov 12 '13 at 15:15
  • $\begingroup$ Paint the equal letters to make distinct. Tie R and V together. You have $6$ "letters". Can be permuted in $6!$ ways. Then R and V can be permuted, so $2\cdot 6!$. Finally, uncolour. This divides by $2!3!$. $\endgroup$ – André Nicolas Nov 12 '13 at 15:23
  • $\begingroup$ Thanks, @AndréNicolas. Do you mean "tie V and S together"? In your first example, you say $\frac{7!-2\cdot6!}{2!\cdot3!}$ but I'm not sure how this ties in with the second example where you don't mention $7!$. $\endgroup$ – Philip Nov 12 '13 at 15:39
  • $\begingroup$ In the second example I was explaining how to just count the "bads." Here is the idea, with people. We have $7$ people, of whom $2$ (V and S) insist on being next to each other. How many ways to line up the people? Put V and S in a bag. We have $6$ "people" ($5$ real and a bag). There are $6!$ ways to line these up in a row. Now untie the bag. Keeping all the rest of the orders, V can be to the left of S or to the right, total $2\cdot 6!$. By the way the "insertion" technique you used in your answer is a good tool too. $\endgroup$ – André Nicolas Nov 12 '13 at 18:23
  • $\begingroup$ And yes, I meant V and S, R was a typo. $\endgroup$ – André Nicolas Nov 12 '13 at 20:52

You can easily solve this backwards.

1) As you calculated there are $420$ ways to arrange those seven letters

2) Calculate how many arrangements there are if $V$ and $S$ is together. Now you have 6 letters (let's take $VS$ or $SV$ as one letter). You get $$\frac{6!}{2!\cdot3!}=60$$ different arrangements, and letters $V$ and $S$ can be arranged in $2$ different ways ($VS$ or $SV$). So there are $60\cdot2=120$ arrangements where $V$ is together with $S$.

3) Subtract this number from all the arrangements and you get the answer: $420-120=300$


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