Checks-check every where, How much I count!

For the $4 \times 4$ case, I'll use the counting principle.

To make the argument simpler, I'll consider $4 \times 4$ $0-1$ matrices $M$ whose row and column sums all equal $2$ . (I.e., $M_{ij} = 1$ if and only if there is a red square in the $i$ th row and $j$ th column.)

There are ${ 4 \choose 2}$ ways to place two red squares in the first row. Let $i_1, i_2$ be such that $M_{1 i_1} = 1 = M_{1 i_2}$ . Consider a fixed value of $i_1$ . There are $3$ choices for $j$ such that $M_{j i_1} = 1$ .

At this point, we split into two cases:

1) $M_{j i_2} = 1$ . If this is the case, then we have two rows and two columns that have their red squares, so the other four reds are $\{ M_{kl}: k \neq 1, j , l \neq i_1, i_2\}$ . Thus there is one matrix in this case.

2) $M_{j i_2} = 0.$ In this case there are two choices for $l \neq i_1, i_2$ such that M_{jl} = 1, and for each of those two choices there are two choices for $k$ such that $M_{k i_2} = 1$ . For each of these four choices, there is exactly one way to complete the matrix to get the desired row and column sums. Thus there are exactly four matrices in this case.

Thus the total number of such matrices is ${4 \choose 2} * 3 * (1 + 4) = 90$

I solved the problem in 3 steps

• 1)No. of ways to fill the first row is $^4C_2$

• 2)No. of ways to fill the first column is $^3C_2$

{as the first block of first column is already filled in step (1) }

• 3)No. of ways to fill second row is $^3C_2$

Now from these 3 methods,

- - One will lead to complete filled board [ when the colours in first row and second row match ] { 1× 1 }

( because there can be only 2 blocks of same colour in a line )

- - Other two will lead to two cases each [ as on filling the opposite colours, we would go to the leftmost empty block of 3rd row. There we will have 2 colours to fill in, so 2 ways ] { 2×2 }

Finally we get the number of ways of filling the board equal to

(1) × (2) × (3)

= $^4C_2$ × $^3C_2$ × ( 1×1 + 2×2 )

= 6 × 3 × 5

= 90