Question

A Tic-tac-toe board has 9 spaces. The first player fills one of the spaces with an X, then the second player fills a space with an O. The players continue to alternate filling a space with their respective symbol until no empty spaces remain on the board. How many different arrangements of X’s and O’s are possible at the end of the game? You must fully justify your answer.

Answer #1

There are 9 spaces to fill

First player started with a X , then the second player started with O

As the number of spaces are odd the filling process will be ended by the first player with a X

So, there will be 5 X's and 4 O's among 9 spaces

9 symbols can be arranged in 9! ways

Among these 9 symbols 5 symbols are same that is X and 4 symbols are same that is O

So, these 9 symbols can be arranged in 9!/(5! 4!) Ways where 5! is is in the denominator for 5 same symbol X and 4! is in the denominator for 4 same symbol O.

So, number of all possible arrangements,

=9!/(5! 4!)

=126

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