Try this beautiful Combinatorics Problem based on colour from integer from Prmo-2018.

## Colour Problem- PRMO 2018- Problem 27

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?

,

- \(28\)
- \(90\)
- \(32\)
- \(16\)
- \(27\)

**Key Concepts**

Chessboard

Combinatorics

Probability

## Suggested Book | Source | Answer

#### Suggested Reading

Pre College Mathematics

#### Source of the problem

Prmo-2018, Problem-27

#### Check the answer here, but try the problem first

\(90\)

## Try with Hints

#### First Hint

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…………..

#### Second Hint

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

#### Third Hint

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

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