Grid game!

We have five squares to choose:

The first square to choose has $100$ ways.

Now, consider a randomly chosen square. The second one should not belong to first's row and column, so we can neglect from the $100$ ways, $19$ ways (The shaded squares), Having $(100-19)$ ways for the second.

For the third square to be chosen, we should neglect the shaded rows and columns of each of the two previously chosen squares, but taking into consideration that the shaded areas will meet at $2$ squares, thus, leaving us with $(100-19-19+2)$ ways for the third.

The walkthrough now is obvious: The shaded rows and columns of the third chosen square will meet the other shadings in $4$ squares, leaving us with $(100-19-19+2-19+4)$ ways for the fourth.

The shadings of the fourth chosen square will meet the other shadings in $6$ squares, leaving us with $(100-19-19+2-19+4-19+6)$ ways for the Fifth.

Therefore the number of ways to choose the five squares is: $(100)(81)(64)(49)(36) = 914457600$ .

But this number of ways counts repeated possibilities, thus divide by $5!$ to get the correct answer:

$914457600/5! = \boxed{7620480}$

The goal of the solution is to find how many sets of 5 points $A(x_1,y_1),B(x_2,y_2),C(x_3,y_3),D(x_4,y_4)$ and $E(x_5,y_5)$ there are, which satisfy $x_i<x_j$ , $y_i \neq y_j$ and $1 \leq x_i,y_i \leq 10$ for all integers $1 \leq i<j \leq 5$ .

How many ways to choose $x-$ values are there? It's $C^5_{10}$ .

And how many ways to choose $y-$ values are there? Careful! It's $P^5_{10}$ , not $C^5_{10}$ . 'Cause for $y$ , group $(1;2;3;4;5)$ is different from group $(2;3;4;5;1)$ , but for $x$ , it's the same.

Number of ways: $C^5_{10} \times P^5_{10} =\boxed{7620480}$ .

Note: Generalising, the number of ways to choose $k$ points in a $m \times n$ grid such that no $2$ squares are in same row or column, and $k \leq \text{min(m,n)}$ is $C^k_{m} \times P^k_{n}$

Ans is 100 * 81 * 64 * 49 * 36 / 5!