Question

Six $X^{\prime }s$ have to be placed in the square of a given figure such that each row contains at least one $X$. Then, number of ways of doing so are

• A. $28$
• B. $26$
• C. $6!$
• D. $8!\cdot 6!$

Answer 1

Answer: (B)

As all $X^{\prime }s$ are identical, In the problem, we need to select $6$ squares from $8$, so that no row remains empty.
This can be done as follows :

$R_1$	$R_2$	$R_3$	Number of selections
1	4	1	${}^2{C}_1\cdot {}^4{C}_4\cdot {}^2{C}_1=4$
1	3	2	${}^2{C}_1\cdot {}^4{C}_3\cdot {}^2{C}_2=8$
2	3	1	${}^2{C}_2\cdot {}^4{C}_3\cdot {}^2{C}_1=8$
2	2	2	${}^2{C}_2\cdot {}^4{C}_2\cdot {}^2{C}_2=6$

Hence, total no. of ways $=4+8+8+6=26$
Alternative Solution:
Required number of ways = Coefficient of $x^6$ in the product of the series for above cases = Coefficient of $x^6$ in ((number of ways of filling at least one box out of two for first row $R_1)\times$ (at least one box out of $4$ for $2$nd row $R_2)\times$ (at least one out of $2$ for third row $R_3$))
$=$ Coefficient of $x^6$ in $({}^2{C}_0{x}^0+{}^2{C}_1{x}^1+{}^2{C}_2{x}^2-1{)}^2\times ({}^4{C}_0{x}^0$
$+{}^4{C}_1{x}^1+{}^4{C}_2{x}^2+...+{}^4{C}_4{x}^4-1)$
(As $R_1$ and $R_3$ has same ways of filling at least one box)
$=$ Coefficient of $x^6$ in $x^3(2+x{)}^2(4+6x+4x^2+x^3)$
$=$ Coefficient of $x^{6-3}$ in $(4+4x+x^2)(4+6x+4x^2+x^3)$
$=$ Coefficient of $x^3$ in $(16+4x^3+16x^3+6x^3+...)$
$=4+16+6=26$
Short Cut Method :
Total number of squares $=8$
Number of places to be filled $=6$
$\therefore$ Number of ways to do so $={}^8{C}_6$
$={}^8{C}_2=28$
These $28$ cases also hold those two cases in which either $R_1$, or $R_3$ may not have any $X$, so exclude these two cases from total $28$ ways.
$\therefore$ Required number of ways $=28-2=26$