More tic-tac-toe!

We first count the number of drawing positions.

This is a drawing position with $4$ rotations. Assume the blank spaces are $X$ 's.

This is a drawing position with $4$ rotations. Assume the blank spaces are $X$ 's.

This is a drawing position with $4$ rotations and $2$ reflections, thus $8$ permutations. Assume the blank spaces are $X$ 's.

Therefore, the total number of drawing positions are $8+4+4 = 16$ . But, notice that in any given position, $O$ can choose any of the $5$ $O$ 's to put in the grid first. Similarly, $X$ can put his next $X$ in $4$ ways, and so, on, so the number of ways to acquire any given position is $5!4! = 2880$ .

But, there are $16$ positions, so the total number of ways to play a drawing game is $16 \cdot 2880 = \boxed{46080}$ .

Relevant wiki: Permutations - Problem Solving

Having a draw in the game of tic-tac-toe means that our 3×3 grid is filled with 5 circles and 4 crosses, in a way that there aren't any 3 crosses or circles that form a row, a column or a diagonal.

There are 9! ways to fill a 3×3 grid with 5 circles and 4 crosses in total.

As there are 3 rows, 3 columns and 2 diagonals (3 + 3 + 2 = 8 key triplets altogether), the number of cases when we have (at least) one of these key triplets

a) made out of crosses is:

$8 ×$ ${6} \choose {1}$ $= 48$

b) made out of circles is:

$8 ×$ ${6} \choose {2}$ $- 9 - 12 - 1 = 98$ , since

there are 8 key triplets, then we have to choose 2 places out of 6 for the remaining two circles and deduct those ways, which we counted twice, as 5 circles can form:

• a row and a column in 9 cases

• a row/column and a diagonal in 6 + 6 = 12 cases

• both diagonals in 1 case,

c) both (double counted):

This can only happen, if we have either two rows or two columns made out of rows/crosses (one each). This can be achieved in ${3} \choose {2}$ $× 2 × 2 = 12$ ways, and we can choose the the positions of the remaining 2 circles out of 3 places.

The number of relevant cases:

$12 ×$ ${3} \choose {2}$ $= 36$

Therefore, we have

98 + 48 - 36 = 110 cases altogether.

As we can occupy the 5 circle / 4 cross positions in any order, the total number of ways to fill the 3 × 3 grid with five circles and four crosses (without considering, whether it is a valid tic-tac-toe game or not) is:

110 × 5! × 4! = 316800 ,

and the number of tic-tac-toe games ending in a draw is:

$9! - 316800 = \boxed {46080}$

good question! got 28 cases where O didn't win without thinking whether x wins or not. X can only win when it is on a diagonal.So total 12 cases where X wins on the diagonal and O doesnt win.Hence total cases will 28-12=16.Then multiply it by 5!4!=2880.So total ways=16*2880=46080