2005 AMC 10A Problem 9

The X tiles all look the same and the O tiles all look the same. So a row is decided completely by where the two O's go. The X's fill the other three spots.

1. Count all possible rows. Choose $2$ of the $5$ spots for the O's. List them by the first O's spot:
- first O in spot 1: second in 2, 3, 4 or 5, giving $4$ ways;
- first O in spot 2: second in 3, 4 or 5, giving $3$ ways;
- first O in spot 3: $2$ ways;
- first O in spot 4: $1$ way.

Total: $4 + 3 + 2 + 1 = 10$ rows. (This is $\binom{5}{2} = 10$.)
2. Are they equally likely? Yes. The tiles are placed at random, so no pattern is favored.
3. Count the good rows. $XOXOX$ has O's in spots 2 and 4. That is exactly $1$ of the $10$.

So the probability is $\frac{1}{10}$.

The answer is $\boxed{\textbf{(B)}\ \frac{1}{10}}$.

When some objects are identical, count the different-looking rows, not every shuffle of the tiles. Picking the spots for the smaller group (the two O's) is the quickest way. If you prefer to treat all five tiles as different, that also works, as long as you count the same way on the top and the bottom of the fraction: $\frac{12}{120}$ is still $\frac{1}{10}$.

Place the tiles one at a time, left to right.

• First tile is X: $3$ of $5$ tiles are X, so $\frac{3}{5}$.
• Second is O: $2$ of the $4$ left are O, so $\frac{2}{4}$.
• Third is X: $2$ of the $3$ left are X, so $\frac{2}{3}$.
• Fourth is O: $1$ of the $2$ left is O, so $\frac{1}{2}$.
• The last tile must be X.

Multiply: $\frac{3}{5} \cdot \frac{2}{4} \cdot \frac{2}{3} \cdot \frac{1}{2} = \frac{12}{120} = \frac{1}{10}$.