Question

Number of ways of arranging $5$ different objects in the squares of given figure in such a way that no row remains empty and one square can't have more then one object.

• A. $44\times 5!$
• B. $48\times 5!$
• C. $36\times 5!$
• D. $40\times 5!$

Answer 1

The correct option is A $44\times 5!$

Following cases will be there:
We are given $5$ different objects

Row No.	Case $1$	case $2$	case $3$	case $4$
$1^{st}$	$3$	$2$	$2$	$1$
$2^{nd}$	$1$	$2$	$1$	$2$
$3^{rd}$	$1$	$1$	$2$	$2$
No. of ways for each case	${(}^4{C}_3{}^2{C}_1{}^2{C}_1)\times 5!$ $=16\times 5!$	${(}^4{C}_2{}^2{C}_2{}^2{C}_1)\times 5!$ $=12\times 5!$	${(}^4{C}_2{}^2{C}_1{}^2{C}_2)\times 5!$ $=12\times 5!$	${(}^4{C}_1{}^2{C}_2{}^2{C}_2)\times 5!$ $=4\times 5!$

Hence, required number of ways
$=(16+12+12+4)\times 5!$
$=44\times 5!$