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In how many ways can the letters of the word ARRANGE be arranged such that two R's are never together?

Now total number of words are $\frac{7!}{2!2!}$.

Now words in which R's are always together are $\frac{6!}{2!2!}$. Subtracting these two doesn't give me correct answer. Can you explain why?

Thanks.

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    $\begingroup$ There are $15$ possible positions for the two R such that they are not together (place one and count positions for the other, always on the right to avoid duplicates, or if you prefer there are ${7 \choose 2}=21$ pairs and $6$ for which the R are together). Hence the total number of words is for this case $15\times5!/2=15\times60=900$. You can check that $7!/(2!2!)=1260$ and as Uddeshya Singh wrote below, there are $6!/2!=360$ words such that the two R are together, and you get the sum right: $360+900=1260$. $\endgroup$ Apr 11, 2017 at 6:11

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In your second case. It will be just $\frac{6!}{2!}$. Because your are taking 'RR' as a single unit which can be adjusted only one way

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  • $\begingroup$ huh ? let me for time being take as R' and R''. won't i get two copied if i don't divide by 2. as an example consider ANGER'R''A and ANGER''R'A. $\endgroup$
    – J. Deff
    Apr 11, 2017 at 5:52
  • $\begingroup$ Won't they both mean the same thing? @J.Deff . We are talking about permutating two IDENTICAL OBJECTS together. it just makes it $\frac{2!}{2!}$ =1!? $\endgroup$ Apr 11, 2017 at 5:53
  • $\begingroup$ they are same word, but two copies $\endgroup$
    – J. Deff
    Apr 11, 2017 at 5:54
  • $\begingroup$ Tell me why would you like to count EGG and EGG as two different words when even the dictionary says they are the same. Basically in this case, we have 6 things to align. "A"x2 , "RR","N","G","E". Thus it makes it $\frac{6!}{2!}*\frac{2!}{2!}$. @J.Deff $\endgroup$ Apr 11, 2017 at 5:58
  • $\begingroup$ then why we have divided by 2, in case of A's $\endgroup$
    – J. Deff
    Apr 11, 2017 at 6:00

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