Posted by **Robin** on Thursday, January 28, 2010 at 12:21pm.

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?

- 8th math -
**drwls**, Thursday, January 28, 2010 at 12:49pm
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

- 8th math -
**drwls**, Thursday, January 28, 2010 at 1:23pm
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.

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