Two players play a game in which they alternate calling out positive integers ≤ N, according to:

The first player must always call out odd numbers.

The second player must always call out even numbers.

Each player must call out a number greater than the previously called number (except, obviously, the very first time).

The player who cannot call out a number loses.

How many different possible games are there? And, if we count a turn each time a player calls out a number, how many different K-turns games are there?

Note: the game is not very fun to play (why?) but the puzzles are interesting!

The number of different K-turn games is related to Pascal's
triangle.

I may have formulated this incorrectly, but here goes....

Such that K > 0 and K <= N, the number of different K-turn games is equal the summation from 0 to N (where N > 0) for
(([(N+1)/2]-1)! / (K-1)!*(([(N+1)/2]-K)!)/2

[ ] is used for the floor function, here

Edited on April 15, 2007, 3:54 pm

Posted by Dej Mar on 2007-04-14 16:06:04