Azar and Carl play a game of tic-tac-toe. Azar places an in $X$ one of the boxes in a $3$-by-$3$ array of boxes, then Carl places an $O$ in one of the remaining boxes. After that, Azar places an $X$ in one of the remaining boxes, and so on until all boxes are filled or one of the players has of their symbols in a row—horizontal, vertical, or diagonal—whichever comes first, in which case that player wins the game. Suppose the players make their moves at random, rather than trying to follow a rational strategy, and that Carl wins the game when he places his third $O$. How many ways can the board look after the game is over?
$\textbf{(A) } 36 \qquad\textbf{(B) } 112 \qquad\textbf{(C) } 120 \qquad\textbf{(D) } 148 \qquad\textbf{(E) } 160$
We need to find out the number of configurations with 3 $O$ and 3 $X$ with 3 $O$ in a row, and 3 $X$ not in a row.
$\textbf{Case 1}$: 3 $O$ are in a horizontal row or a vertical row.
Step 1: We determine the row that 3 $O$ occupy.
The number of ways is 6.
Step 2: We determine the configuration of 3 $X$.
The number of ways is $\binom{6}{3} - 2 = 18$.
In this case, following from the rule of product, the number of ways is $6 \cdot 18 = 108$.
$\textbf{Case 2}$: 3 $O$ are in a diagonal row.
Step 1: We determine the row that 3 $O$ occupy.
The number of ways is 2.
Step 2: We determine the configuration of 3 $X$.
The number of ways is $\binom{6}{3}  = 20$.
In this case, following from the rule of product, the number of ways is $2 \cdot 20 = 40$.
Putting all cases together, the total number of ways is $108 + 40 = 148$.
Therefore, the answer is $\boxed{\textbf{(D) }148}$.
~Steven Chen (www.professorchenedu.com)