Question

Consider the letters of the word ${}^{\prime }ARRANG{E}^{\prime }$. Then the possible number of words

• A. when $2R^{\prime }$s are not together is $\frac{5!}{2!}\times {}^6{P}_2=900$
• B. when two $A^{\prime }$s are together but not two $R$'s is $240$
• C. when the vowels are together is $\frac{5!3!}{2!2!}$
• D. when both $A^{\prime }$s are together and both $R^{\prime }$s are together is $120$

Answer 1

Answer: (B), (C), (D)

The correct options are
B when two $A^{\prime }$s are together but not two $R$'s is $240$
C when the vowels are together is $\frac{5!3!}{2!2!}$
D when both $A^{\prime }$s are together and both $R^{\prime }$s are together is $120$

When $2R^{\prime }$s are not together.
Arrange the letters ${}^{\prime }A,A,N,G,E$ in alternate manner, such that there will be $6$ spaces left, so that can be filled by 2$R^{\prime }$s in${}^6{C}_2$ ways
So the total possible words $=\frac{5!}{2!}\times {}^6{C}_2=900$

When two $A^{\prime }$s are together but not two $R^{\prime }$s.
The number of words in which both $A$'s are together
$=\frac{6!}{2!}$ $[$consider both the $A^{\prime }$s as one unit$]$
$=360$
The number of words in which both $A^{\prime }$s and both $R^{\prime }$s are together is $5!=120$
$[$consider both the $A^{\prime }$s as one unit and both the $R^{\prime }$s as one unit$]$
Therefore, the number of words in which both $A^{\prime }$s are together but the two $R^{\prime }$s are not together is $360-120=240$

When the vowels are together.
Make one group of all vowels.
The total ways of arrangement is $=\frac{5!\times \frac{3!}{2!}}{2!}$

When both $A^{\prime }$s are together and both $R^{\prime }$s are together.
Make one group $2A^{\prime }$s and another group of $2R^{\prime }$s.
The total ways of arrangement is $=5!=120$