posted by Robin on .
How many ways can seven basketball players of different heights line up in a single row so no player is standing between two players taller then herself?
Two would be ascending or descending order.
Twelve more would be ascending or descending order, with any one of the not-tallest going to the other end.
I can think of four others that involve flipping and inverting more than one to the opposite side. Maybe you can think of others. Add them up
By seeing how many arrangements I could make with N players, I found that for N = 1, 2, 3, 4, and 5 players, the number of arrangements is 2^(N-1).
If this trend continues, the number of possible arrangements of N = 7 players is 2^6 = 64. Figuring it out with numbers (1 through 7) is easier than doing it with "straws" of different length.