3. $5 \mathrm{Xs}$ and 4 Os are arranged in the below grid such that each number is covered by either an X or an O. There are a total of 126 different ways that the Xs and Os can be placed. Of these 126 ways, how many of them contain a line of 3 Os and no line of 3 Xs?
A line of 3 in a row can be a horizontal line, a vertical line, or one of the diagonal lines $1-5-9$ or $7-5-3$.
Solution: If we have a horizontal (or vertical) line of all Os, then since there are $5 \mathrm{Xs}$ for the other two lines, there must be a horizontal (or vertical) line of all Xs. Thus, our line of 3 Os must be a diagonal.

When one of the diagonal lines is all Os, then no other line can be all Xs, since each diagonal line intersects all other lines. Thus, each configuration with one diagonal line of Os is a desired solution.
When we have the diagonal line $1-5-9$, there are 6 places that the last $O$ could be: $2,3,4,6,7,8$. Each of these will give a valid solution. Similarly, we have 6 solutions when we have the diagonal line $7-5-3$.
It is not possible for both diagonal lines to have only Os, since there are only 4 Os, thus we have not counted the same configuration twice. Thus 12 of the 126 ways contain a line of 3 Os and no line of $3 \mathrm{Xs}$.