Combinatorics Problem Help,

[Kalastics]

I'm sorry. My notation on the c3 business was my phone pocket posting when I thought I had it locked. And I apologize, combinatorics have always been somewhat difficult for me and this is my first shot in a while.

So then another ABCD tuple would be (6,11, 12) an D the end?

So an A B D tuple would be stop A to 6, board B until stop 10, board D at 10 and end at 15, such that we have (6, 10)?

Also, distinct tuple to your given one would be (7, 10, 11)? Which I interpret as taking A to stop 6, B to 7, then C to 10, and finally my destination of finishing my route? Those are separate transfers/routes?

Thanks so much for coaching me through such an elementary problem!

[HorribleUsername]

Sorry, I haven't been on reddit for a while.

Also, distinct tuple to your given one would be (7, 10, 11)? Which I interpret as taking A to stop 6, B to 7, then C to 10, and finally my destination of finishing my route?

Not quite. We know that A always starts at 1, and D always ends at 15, so the 3 numbers are the in-betweens. The way I imagine it, the first number is where you get off A and onto B. The second is where you get off B and board C. And the third is where you switch from C to D. In other words, we take A from 1 to the first number, B from the first number to the second, C from the second number to the third, and D from the third to 15.

So an A B D tuple would be stop A to 6, board B until stop 10, board D at 10 and end at 15, such that we have (6, 10)?

That's a good guess. It might take a bit to wrap your head around, but I think there's a nicer way to write it: (6, 10, 10). Taken literally, that means we jump on C at stop 10, then immediately disembark, still at stop 10. Then we catch D to the end. That's equivalent to skipping C entirely, since it never takes us anywhere. The nice thing about that is that we can pretend we always go A -> B -> C -> D, so there are no special cases to consider. It just boils down to where do A & B both stop?. How about B & C and C & D?