A visual representation of a flat grid or chart, denoted by x-y coordinates with (0,0) at the center. The grid should be designed such that eight distinct paths can be traced from the center back to itself, as if an ant moves in the directions of North, South, East, and West, one unit at a time. The paths are visualized as lines in 8 different colors and an ant figure is illustrated at the start (0,0) point. The image excludes any form of text.

An ant starts at position (0, 0) and walks north, east, south or west, one unit at a time.

How many di�erent paths of length 8 units finish at (0, 0)?

move 1 unit East: 8!/3!3!=1120

move 2 units East: 8!/2!2!2!2!=2520
move 3 units East: 8!/3!3!=1120
move 4 units East: 8!/4!4!=70
move 0 unit East: 8!/4!4!=70
Total paths: 4900

Note: if ant moves 1 unit East, must move back 1 unit West.

Take a look at the following article, and see what you can come up with:

https://betterexplained.com/articles/navigate-a-grid-using-combinations-and-permutations/

To find the number of different paths of length 8 units that finish at (0, 0), we can use the concept of combinations.

Let's visualize the problem on a coordinate plane. The ant starts at position (0, 0), and after 8 units of movement, it also ends up at (0, 0). Since the ant can only move one unit at a time in north, east, south, and west directions, there are only two possible movements for each unit:

1. Move north and then south.
2. Move east and then west.

Considering this, we can calculate the number of different paths using combinations. In this case, we need to choose 4 units to move north and 4 units to move east.

The formula for combinations is given by:

nCr = n! / (r!(n-r)!)

where n represents the total number of units and r represents the number of units to choose for a particular movement.

For our case, n = 8 and r = 4.

Plugging the values into the formula:

8C4 = 8! / (4!(8-4)!)
= 8! / (4!4!)

Now, let's calculate the value using the factorial function.

8! = 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 = 40320
4! = 4 x 3 x 2 x 1 = 24

Plugging the values back into the formula:

8C4 = 40320 / (24 x 24)
= 40320 / 576
= 70

Therefore, the number of different paths of length 8 units that finish at (0, 0) is 70.

To solve this problem, we can use a simple combinatorics approach.

For each step, the ant has four possible directions to choose from: north, east, south, or west. Since the ant is starting at (0, 0) and wants to finish at (0, 0) after 8 units, the ant must take an equal number of north and south steps as well as an equal number of east and west steps.

Let's consider the number of north/south steps the ant takes. Since the ant must take an equal number of north and south steps, the number of steps in each direction can range from 0 to 4. We can represent these steps as N or S. For example, NNSS means two north steps followed by two south steps.

Similarly, the number of east/west steps the ant takes must also be equal. We can represent these steps as E or W.

Therefore, our task is to find all possible combinations of N's and S's as well as E's and W's that both add up to 8.

We can use a combinatorics formula called the binomial coefficient to find this. The formula is given as:

C(n, k) = n! / (k!(n-k)!)

Where n is the total number of steps (in this case, 8), and k is the number of steps in one direction (north/south or east/west).

Now let's calculate the number of different paths:

For the north/south steps:
C(8, 0) = 8! / (0!(8-0)!) = 1
C(8, 1) = 8! / (1!(8-1)!) = 8
C(8, 2) = 8! / (2!(8-2)!) = 28
C(8, 3) = 8! / (3!(8-3)!) = 56
C(8, 4) = 8! / (4!(8-4)!) = 70

For the east/west steps, we have the same number of combinations:

C(8, 0) = 8! / (0!(8-0)!) = 1
C(8, 1) = 8! / (1!(8-1)!) = 8
C(8, 2) = 8! / (2!(8-2)!) = 28
C(8, 3) = 8! / (3!(8-3)!) = 56
C(8, 4) = 8! / (4!(8-4)!) = 70

Finally, we can calculate the number of different paths by multiplying the number of combinations for north/south steps by the number of combinations for east/west steps:

Total number of paths = (1 * 1) + (8 * 8) + (28 * 28) + (56 * 56) + (70 * 70) = 1 + 64 + 784 + 3136 + 4900 = 7885.

Therefore, there are a total of 7885 different paths of length 8 units that finish at (0, 0).

Ah, the adventures of our intrepid ant! Let's see, the ant needs to take 8 steps to complete its journey back to the starting point. Now, every step can either be north, east, south, or west.

Think of it this way - each step is like a multiple-choice question with four options. And since the ant needs to take 8 steps in total, it's like solving an 8-question multiple-choice quiz! The number of possible paths is basically the number of ways the ant can answer these quiz questions.

So, for each step, the ant has 4 options. Since there are 8 steps in total, we need to multiply the number of options for each step, which is 4, a total of 8 times.

Mathematically, this can be expressed as 4^8, which is equal to 65,536.

Therefore, there are 65,536 different paths that our ant friend can take to end up back at our starting point, (0, 0). I hope the ant finds its way back without getting too antsy!