2004 AMC 10B Problem 1

1. First count the rows. It is tempting to do $22 - 12 = 10$, but try a small case to check that idea: rows $1$ through $3$ are rows $1, 2, 3$, which is $3$ rows, while $3 - 1 = 2$. Subtracting misses one row, because both end rows are included.
2. So rows $12$ through $22$ are $22 - 12 + 1 = 11$ rows. You can also see this by listing them: $12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22$. That is $11$ labels.
3. Each row has $33$ seats, so the club gets
$$11 \times 33 = 363$$
seats.

The answer is $\boxed{\textbf{(C)}\ 363}$.

When a problem says "from $12$ through $22$," both ends are in the list, so the count is (last minus first) plus one. If you are ever unsure, test the rule on a tiny list you can count by hand, like $1$ through $3$. Choice (B) $330$ is exactly the answer you get by forgetting the plus one.

First count the rows. The labels $12, 13, \dots, 22$ form a list of $22 - 12 + 1 = 11$ numbers; one way to see this is to subtract $11$ from every label, which turns the list into $1, 2, \dots, 11$.

Each of these $11$ rows holds $33$ seats, so the club gets
$$11 \cdot 33 = 363$$
seats.

The answer is $\boxed{\textbf{(C)}\ 363}$.