Try this beautiful Combinatorics Problem based on colour from integer from Prmo-2018.
What is the number of ways in which one can colour the square of a $4 \times 4$ chessboard with colours red and blue such that each row as well as each column has exactly two red squares and blue
squares?
,
Chessboard
Combinatorics
Probability
Pre College Mathematics
Prmo-2018, Problem-27
\(90\)
First row can be filled by ${ }^{4} \mathrm{C}_{2}$ ways $=6$ ways.
Case-I
Second row is filled same as first row
$\Rightarrow$
here second row is filled by one way
$3^{\text {rd }}$ row is filled by one way
$4^{\text {th }}$ row is filled by one way
Total ways in Case-I equals to ${ }^{4} \mathrm{C}_{1} \times 1 \times 1 \times 1=6$ ways
now we want to expand the expression and simplify it..............
Case-II $\quad$ Exactly $1$ R & $1$ B is interchanged in second row in comparision to $1^{\text {st }}$ row
$\Rightarrow$
here second row is filled by $2 \times 2$ way
$3^{r d}$ row is filled by two ways
$4^{\text {th }}$ row is filled by one way
$\Rightarrow$
Total ways in Case-II equals to ${ }^{4} \mathrm{C}_{1} \times 2 \times 2 \times 2 \times 1=48$ ways
Case-III $\quad$ Both $\mathrm{R}$ and $\mathrm{B}$ is replaces by other in second row as compared to $1^{\text {st }}$ row
$\Rightarrow$
here second row is filled by 1 way
$3^{r d}$ row is filled by $4 \choose 2 $ ways
$\Rightarrow \quad$ Total ways in $3^{\text {th }}$ Case equals to ${ }^{4} \mathrm{C}_{2} \times 1 \times 6 \times 1=36$ ways
$\Rightarrow \quad$ Total ways of all cases equals to 90 ways
Try this beautiful Combinatorics Problem based on colour from integer from Prmo-2018.
What is the number of ways in which one can colour the square of a $4 \times 4$ chessboard with colours red and blue such that each row as well as each column has exactly two red squares and blue
squares?
,
Chessboard
Combinatorics
Probability
Pre College Mathematics
Prmo-2018, Problem-27
\(90\)
First row can be filled by ${ }^{4} \mathrm{C}_{2}$ ways $=6$ ways.
Case-I
Second row is filled same as first row
$\Rightarrow$
here second row is filled by one way
$3^{\text {rd }}$ row is filled by one way
$4^{\text {th }}$ row is filled by one way
Total ways in Case-I equals to ${ }^{4} \mathrm{C}_{1} \times 1 \times 1 \times 1=6$ ways
now we want to expand the expression and simplify it..............
Case-II $\quad$ Exactly $1$ R & $1$ B is interchanged in second row in comparision to $1^{\text {st }}$ row
$\Rightarrow$
here second row is filled by $2 \times 2$ way
$3^{r d}$ row is filled by two ways
$4^{\text {th }}$ row is filled by one way
$\Rightarrow$
Total ways in Case-II equals to ${ }^{4} \mathrm{C}_{1} \times 2 \times 2 \times 2 \times 1=48$ ways
Case-III $\quad$ Both $\mathrm{R}$ and $\mathrm{B}$ is replaces by other in second row as compared to $1^{\text {st }}$ row
$\Rightarrow$
here second row is filled by 1 way
$3^{r d}$ row is filled by $4 \choose 2 $ ways
$\Rightarrow \quad$ Total ways in $3^{\text {th }}$ Case equals to ${ }^{4} \mathrm{C}_{2} \times 1 \times 6 \times 1=36$ ways
$\Rightarrow \quad$ Total ways of all cases equals to 90 ways
