Question

Number of ways in which $8$ people can be arranged in a line if $A$ and $B$ must be next each other and $C$ must be somewhere behind $D$ is equal to

• A. $10080$
• B. $5040$
• C. $5050$
• D. $10100$

Answer 1

The correct option is B $5040$

Now as $A$ and $B$ are next to each other so we assume them as one. So now total people will be $7$.

Now $C$ should be somewhere behind $D$, therefore we need to consider all cases according to positions of $D.$

$\left(i\right)$ When $D$ is at first position $C$ can be placed at any remaining $6$ positions=Total ways to arrange remaining $6$ people$(D$ is at first$)=6!$

$\left(ii\right)$When $D$ is at second position $C$ can be placed only at remaining $5$ positions after $D$, so at first position there can be only $5$ people$=$Total ways will be$=5\times 5!$

$\left(iii\right)$When $D$ is at third position $C$ can be placed only at remaining $4$ positions after $D$, so at first and second there can be only $5\times 4$ choices$=$Total ways will be$=5\times 4\times 4!$

Similarly $\left(iv\right)$ When $D$ is at fourth position, Total ways $=5\times 4\times 3\times 3!$

$\left(v\right)$When $D$ is at fifth position, Total ways$=5\times 4\times 3\times 2\times 2!$

$\left(vi\right)$When $D$ is at sixth position, Total ways$=5\times 4\times 3\times 2\times 1\times 1!$

$\left(vii\right)D$ can't be at seventh position because that's the last position so then $C$ can't be behind $D$.

Now as $A$ and $B$ together can sit in two ways$=AB$ or $BA$

$\therefore$ Total ways$=2\times \left[\right(i)+(ii)+(iii)+(iv)+(v\left)\right]$

$=2\times [6!+5\times 5!+5\times 4\times 4!+5\times 4\times 3\times 3!+5\times 4\times 3\times 2\times 2!+5\times 4\times 3\times 2\times 1\times 1!]$

$=2\times [720+600+480+360+240+120]=2\times 2520=5040$

Hence, $\left(B\right)$