Rectangles and squares

Every rectangle on the grid can be drawn by choosing two points on different vertical and horizontal lines and making them the corners of a rectangular box, for example: Using this, we can calculate the total number of rectangles. To pick the first point, there are 6 horizontal lines and 9 vertical lines, so the total number of places to put the first point is $6\times9$ . For the second point, we can't put it on the same row or column, so there are only $5\times8$ places to put it. After multiplying these together, we need to divide by 4, as each rectangle can be drawn from any of its 4 corners, and so the total number of rectangles is $\frac{6\times9\times5\times8}{4}=540$ .

Now, to find the number of rectangles with a corner at A, we can just let the first point we draw the rectangle from be A and the second point be on some other vertical and horizontal line, so the total number is $5\times8=40$ . Note that we don't need to divide by anything, because all the rectangles are being drawn from the same starting corner.

This means the probability of choosing a rectangle with a corner at A is $\large\frac{40}{540}=\frac{2}{27}$

Consider making a rectangle first and then placing point A. Since the position of A has nothing to do with the rectangle, that is valid. Now the ending of the solution is obvious: there are 54 possible places for A out of which 4 fit the conditions, so the answer is $\frac {4}{54}=\frac {2}{27}$ .

We note that any pair of vertical lines and any pair of horizontal lines in the grid form a rectangle. Since it is a $8 \times 5$ grid, there are 9 vertical lines and 6 horizontal lines, and the total number of possible rectangles can be formed is $N = \dbinom 92 \dbinom 62 = 540$ .

By fixing $A$ , we fix a vertical line and horizontal line through $A$ for the two sides the rectangle, we can only select the remaining vertical line from the remaining 8 vertical lines and the remaining horizontal lines from the remaining 5 horizontal lines. The the number of rectangles with a vertex at $A$ is $N_A = \dbinom 81 \dbinom 51 = 40$ .

Therefore, the probability of rectangles with a vertex at $A$ is $p = \dfrac {N_A}N = \dfrac {40}{540} = \boxed{\dfrac 2{27}}$ .

There are 9x6 = 54 potential vertices to create a rectangle in this box. A rectangle has 4 vertices (a square is a type of rectangle). So the probability that A is one of those is 4/54 = 2/27

Generalization :

For $n\times m$ grid the probability is: $\dfrac{4}{(n+1)(m+1)}$

A rectangle is defined by two pairs of parallel grid lines. There are $\binom{n+1}{2} \cdot \binom{m+1}{2}$ of them in total.

A point $A$ is defined as an intersection of two perpendicular lines. It follows that the number of rectangles with vertex $A$ is equal to $\binom{n+1-1}{1} \cdot \binom{m+1-1}{1} = n\cdot m$ .

The probability is the ratio of these two quantities: $\dfrac{n\cdot m}{\binom{n+1}{2} \cdot \binom{m+1}{2}} = \dfrac{n\cdot m}{\frac{(n+1)n}{2}\cdot \frac{(m+1)m}{2}} = \dfrac{4}{(n+1)(m+1)}$

For a vertex to land on $A$ , both a horizontal line and a vertical line must land on $A$ .

The chance of the first chosen vertical line landing on $A$ is $\frac{1}{9}$ . For the remaining $\frac{8}{9}$ times where the line misses, there is now a $\frac{1}{8}$ chance of the second vertical line landing on $A$ . This gives us a $\frac{1}{9}+\frac{1}{8}\times \frac{8}{9}=\frac{2}{9}$ chance of one of the vertical lines being correct.

Similarly, the chance of a horizontal line landing on $A$ is $\frac{1}{6}+\frac{1}{5}\times \frac{5}{6}=\frac{2}{6}$ .

The chance of both of these conditions being true is therefore $\frac{2}{9}\times \frac{2}{6}=\boxed{\frac{2}{27}}$ .

A random rectangle is selected by apppointing two different horizontal lines, and two different vertical lines. The probability that A is on one of the chosen horizontal lines is $\frac{2}{6}$ , the probability that A is on one of the chosen vertical lines is $\frac{2}{9}$ . The probability that A is on a vertex corresponds to the probability that both above conditions are met, and (since the choices of horizontal and vertical lines are independent) this is just their product: $\frac{2}{6}$ × $\frac{2}{9}$ = $\frac{2}{27}$ .

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#include<stdio.h>

#define W 8
#define H 5
#define Ax 3
#define Ay 2

int main() {
    int count = 0;
    int countA = 0;

    for (int x = 0; x < W; x++) for (int y = 0; y < H; y++)
        for (int xx = x+1; xx <= W; xx++) for (int yy = y+1; yy <= H; yy++) {
            count++;
            if ((x == Ax || xx == Ax) && (y == Ay || yy == Ay)) countA++;
        }

    printf("%d / %d \n", countA, count);
}

Output: 40 / 540 , which we reduce to $\boxed{2/27}$ .

There are 6 horizontal lines where two sides of the rectangle may lie, so $1/3$ may have a vertex on A. Of these $2/9$ verticals will have a vertex on A. So the probability will be $1/3 * 2/9 = 2/27$ .

It doesn't matter the shape or layout of the 4 selected vertices. There are 54 (9 * 6) potential vertex points, and there are always 4 chosen vertices. 4/54 = 2/27.

Here's how i found the answer without actually solving it. I counted the number of possibilities in the sample space (9x6=54) and found the denominator from given options which is a multiple of 54.

I just made a guess, I only know that the total number of rectangles in the figure is 540... So, we need to find the probability of a rectangle having vertex $A$ ,

$\therefore$ , it is obvious that the denominator should be $540$ .. and in the options the only number which is a factor of 540 is 27!

Hence, the answer is $\frac{2}{27}$ !

The total number of rectangles is ${9}\choose{2}$ $\times$ ${6}\choose{2}$ .

The number of rectangles with vertex A can be divided up into 4 sections.

For each section we only have to count the different possible opposite corners:

Section 1: $3$ $\times$ $2$ $=$ $6$ . Section 2: $5$ $\times$ $2$ $=$ $10$ . Section 3: $3$ $\times$ $3$ $=$ $9$ . Section 3: $5$ $\times$ $3$ $=$ $15$ .

Adding those up and dividing them by the to total number of rectangles gives up the answer:

There are $\binom{6}{2} \times \binom{9}{2} = \dfrac{5 \times 6 \times 8 \times 9}{4} =540$ sub-rectangles in the given $8 \times 5$ grid of which $40$ sub-rectangles have A as a vertex. As such, the desired probability is $\dfrac{40}{540} = \dfrac{2}{27}$ .

(2x2)/(9x6)=2/27

For you to be able to draw a rectangle from a grid, you need 2 parallel vertical and horizontal lines. So to get the total number of rectangles you can draw in an 8x5 grid, you can choose 2 from 9 vertical lines and 2 from 6 horizontal lines. Thus, $\binom 92$ x $\binom 62$ = 36 x 15 = 540.

You can separate the 8x5 grids into four quandrants setting point A as the origin. Here you will be able to create 4 smaller grids (3x2, 5x2, 3x3 and 5x3). To make sure that the vertex of the rectangle falls on point A, you can set the vertical and horizontal line already passing at point A. This means that for the 3x2 grid, you only get to choose 1 from the 3 remaining vertical lines and 1 from the 2 remaining vertical lines. Thus, $\binom 31$ x $\binom 21$ = 3 x 2 = 6. Applying to the other grids, you obtain 10, 9 and 15 respectively.

Thus, the probability the rectangle has a vertex at point A is $\frac{6+10+9+15}{540}$ = $\frac{2}{27}$ .

The probability of random rectangle is touching vertex A by any of its vertices is 1/(6x9) = 1/54. Since every rectangle has 4 vertices, the probability is 4/54 => 2/27.

Answer = 40/540

The total number of rectangles with the top left being the first square on the grid is $8 \times 5$ . The total number of rectangles with the top left being the second square on the grid is $7 \times 5$ . This pattern continues, so the total number of rectangles with the top left being on the first row of the grid is $8 \times 5 + 7 \times 5 + 6 \times 5 + ... 1 \times 5 = 5 \times (8+7+6+...+1)$ .

The same logic can be applied to the second row, but the column length is four instead of five, so the total number of rectangles with the top left being on the second row of the grid is $4 \times (8+7+6+...+1)$ . The pattern continues, so the total number of rectangles is $5 \times (8+7+6+...+1) + 4 \times (8+7+6+...+1) + 3 \times (8+7+6+...+1) + ... + 1 \times (8+7+6+...+1) = (5+4+...+1)(8+7+6+..+1) = (\frac{5\times 6}{2} \times \frac{8\times 9}{2} = 15 \times 36$ .

If any square is chosen on the grid, a unique rectangle can be created with $A$ and that square being opposite corners of the rectangle. Therefore the total number of rectangles with point $A$ as a vertex is $8 \times 5 = 40$ .

The total probability, therefore, is $\frac{40}{15 \times 36} = \boxed{\frac{2}{27}}$ .