I cannot prove this......but just fooling around, I noticed the following pattern.....
If we have a square.....there are 2 distinct paths of 2 units in length = 1 1 the first row of Pascal's Triangle {I'm calling the single 1 at the top of the Triangle, Row 0 }
To make this follow a pattern, I will write this as 0 1 1 0 ..... notice that the "middle" two numbers sum to 2
Now....it we have 2 x 2 squares = 4 squares....the number of distinct paths of 4 units in length = 6 {prove this for yourself}
Notice that this is the sum of the middle two numbers of the the third row of Pascal's triangle 1 3 3 1
And if we have 3 x 3 squares = 9 squares......the number of distinct paths of 6 units in length = 20
Notice that this is the sum of the middle two numbers of the 5th row of Pascal's Triangle 1 5 10 10 5 1
This makes me wonder that if we have 4 x 4 squares = 16 squares.....that the number of distinct paths of 12 units in length will equal 70.....in other words the sum of the middle two numbers in the 7th row of Pascal's Triangle
1 7 21 35 35 21 7 1 (????? )
The pattern for the number of distinct paths in an n x n arrangement appears to be 2 * C(2n-1, n) for n > 1
Can someone prove/disprove this???
