Actually, for the combinations, you can look at Pascal's triangle and trace it out. For the example from earlier, recall that \(\binom80\) is the \(0\)th number in the \(8\)th row. Doing this for each combination, you can see that it traces out a diagonal. That's probably the thought process I would use for converting the Pascal's Triangle into combinations

Yeah! After it equals \(1\) again, every combination after that equals \(0\), because if you have a combination where the numerator is smaller than the denominator, it equals \(0\) (this goes back to the 3 choose 9 example I was talking about. It doesn't make sense, right?) Right! Exactly. That's why the example I used was actually \(F_9\) since the first combination was \(\binom80\).And yes, great catch. \(\binom80=1\neq0\), I'll go and fix that now ðŸ˜… thank you!!

Great work!

]]>We should let @debbie and @po edit lol ]]>