I want to find out the number of ways to reach a Nth number without having the scores tied.
Example to reach 2-2 I have 2 ways to reach it without having the scores tied and 6 ways to reach it tied 0,0 > 0,1 > 0,2 > 1,2 > 2,2 0,0 > 1,0 > 2,0 > 2,1 > 2,2 Whilst 0,0 > 1,0 > 1,1 is not allowed because it is a tie
What I have already found out is that we can find the amount ways to reach a number in pascal's triangle by this sequence {n \choose r} = t(\frac{\n!}{r!(n-1)!}
What I want to know is how is the sequence related to pascal's triangle and how we could find any Nth number
Sequence number on OEIS A284016 if it helps, I am too confused by it