number of ways of arranging the letters of the word ARRANGE in which two R's are always together

Question

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.

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: $360+900=1260$. — Jean-Claude Arbaut, Apr 11, 2017 at 6:11

The Dead Legend · Accepted Answer · 2017-04-11 05:43:43Z

3

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

answered Apr 11, 2017 at 5:43

The Dead Legend

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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!? $\endgroup$
– The Dead Legend
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 $\endgroup$
– The Dead Legend
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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