Question

In how many ways can be the letters of the word ARRANGE be arranged so that
(i) The two $R^{\prime }s$ are never together
(ii) The two $A^{\prime }s$ are together but not two $R^{\prime }s$
(iii) Neither two $A^{\prime }s$ nor the two $R^{\prime }s$ are together.

Answer 1

There are 7 letters $2A^{\prime }s,2R^{\prime }s$ and 3 different.
(i) Two $R^{\prime }s$ and thus we have to arrange remaining 5 letters in which $2A^{\prime }s$ are alike and we can have
$\frac{5!}{2!}=5\times 4\times 3=60$ ways.
After arranging these five letters in a line we have 6 gaps in which 2 different words can be placed in ${}^6{P}_2$ ways. But these two $R^s$ are alike and hence the number of ways will be.
$\frac{1}{2!}{\cdot }^6{P}_2=\frac{1}{2!}\cdot \frac{6!}{4!}=15$.
Hence the number of ways when $R^s$ are never together is
$60\times 15=900$, by Fundamental $Th$.
(ii) The two $A^s$ ate together but npt two $R^s$.
Let us ignore the two $R^s$ and thus we have $7-2=5$ letters.
Treat the $a{A}^s$ as one letter (AA)NGE thus we have 4 letters which can be arranged in
$4!\frac{2!}{2!}=24$ ways.
$\times AA\times N\times G\times E\times$
between these 4 letter ($A^s$ together) we have 5 gaps in which 2 different letters can be arranged in ${}^5{P}_2$ways.
But here $2R^s$ are alike and hence the number of ways will be
${\frac{1}{2!}}^5{P}_2=\frac{1}{2}\cdot \frac{5!}{3!}=10$ ways.
Hence the number of words when $A^s$ are together and $R^s$ are not together is $24\times 10=240$ by fundamental theorem.
(iii) Neither $2A^s$ not $2R^s$ are together.
From part (i), $2R^s$ are never together in 900 ways.
This includes the ways when $2A^s$ may be together and may not be together.
From part (ii) no. ways that $2A^s$ are together but not $2R^s=240$.
hence the number of ways when neither $2A^s$ not $2R^s$ are together is $900-240=660$.