2005 AMC 10B Problem 21

Both $p$ and $q$ are "good draws divided by all possible draws." The bottom is the same for both, so it cancels in $q/p$. We only need to count the good draws for each.

Count draws as groups of $4$ slips, where order does not matter.

1. Good draws for $p$ (all four slips the same). Pick the number: $10$ ways. There are exactly $4$ slips with that number, and we take all of them: $1$ way. Total: $10$.
2. Good draws for $q$ (two of $a$, two of $b$).
- Pick the two numbers. Picking $3$ and $7$ is the same as picking $7$ and $3$, so there are $\frac{10 \cdot 9}{2} = 45$ ways.
- For the first number, pick $2$ of its $4$ slips. With slips W, X, Y, Z the options are WX, WY, WZ, XY, XZ, YZ: $6$ ways.
- Same for the second number: $6$ ways.
- Total: $45 \cdot 6 \cdot 6 = 1620$.
3. Ratio.
$$\frac{q}{p} = \frac{1620}{10} = 162.$$

The answer is $\boxed{\textbf{(A)}\ 162}$.

When you compare two probabilities from the same experiment, the "all possible outcomes" part is the same, so you never need to compute it. Just count the good outcomes for each. The recipe "choose which numbers, then choose which slips of each number" works for any card-hand style count; the key detail is that the two numbers form an unordered pair.

You can also count draws in order (first slip, second slip, and so on), as long as you do it for both $p$ and $q$. Four of a kind: $10$ numbers, and the $4$ slips can come out in $4 \cdot 3 \cdot 2 \cdot 1 = 24$ orders, giving $240$. Two pairs: $45$ choices of numbers, $6$ choices of which two positions (of the four draws) get the number $a$, and $4 \cdot 3$ orders of slips for each number, giving $45 \cdot 6 \cdot 12 \cdot 12 = 38880$. The ratio $38880 / 240 = 162$ again.