Question

How many arrangements of the word $ARRANGE$ can be made
$\left(i\right)$ If the two $R^{\prime }s$ are not allowed to come together, let it be k?
$\left(ii\right)$ If neither the two $R^{\prime }s$ nor the two $A^{\prime }s$ are allowed to come together, let it be m.
Find the sum of digits of m+k ?

Answer 1

$ARRANGE$ can be made
$\left(i\right)$ If the two $R^{\prime }s$ are not allowed to come together.

Let it be $k$,
then $k=$ number of all possible arrangement - number of arrangement in which two R are together
$k=\frac{7!}{2!\times 2!}$ $-\frac{6!}{2!}$$=900$

$\left(ii\right)$ If neither the two $R^{\prime }s$ nor the two $A^{\prime }s$ are allowed to come together. Let it be $m$.
$k=$ number of all possible arrangement $-$ number of arrangement in which two R and two A are together
$m=\frac{7!}{2!\times 2!}$ $-5!$ $=1140$
Now $m+k=2040$
$\therefore$ Sum of the digits is $6$.