1. The problem asks: In how many ways can you choose 2 cards out of 3 playing cards? 2. This is a combination problem where order does not matter. 3. The formula for combinations is $$\binom{n}{k} = \frac{n!}{k!(n-k)!}$$ where $n$ is the total number of items, and $k$ is the number of items to choose. 4. Here, $n=3$ and $k=2$, so: $$\binom{3}{2} = \frac{3!}{2!(3-2)!} = \frac{3 \times 2 \times 1}{2 \times 1 \times 1} = \frac{6}{2} = 3$$ 5. Therefore, there are 3 ways to choose 2 cards out of 3. Final answer: 3