We are tasked with determining how many 8-card hands drawn from a standard 52-card deck contain at least 2 queens.

Steps to solve the problem:

1. Total number of 8-card hands:
   The number of ways to select 8 cards from a deck of 52 cards (without any restrictions) is given by the combination formula: $\binom{52}{8}$ where $\binom{n}{r}$ is the combination formula, $n$ is the total number of items, and $r$ is the number of items to choose.

2. Hands with less than 2 queens: We need to calculate how many hands have fewer than 2 queens, i.e., either 0 or 1 queen.

   • Hands with 0 queens:
     There are 4 queens in the deck. If we exclude the queens, there are 48 non-queen cards remaining. We can select all 8 cards from these 48 non-queen cards: $\binom{48}{8}$

   • Hands with exactly 1 queen:
     We first select 1 queen from the 4 available queens and then select the remaining 7 cards from the 48 non-queen cards: $\binom{4}{1} \times \binom{48}{7}$

3. Hands with at least 2 queens: To find the number of hands with at least 2 queens, we subtract the number of hands with fewer than 2 queens from the total number of hands: $\text{At least 2 queens} = \binom{52}{8} - \left(\binom{48}{8} + \binom{4}{1} \times \binom{48}{7}\right)$

Final Calculation:

Now, let's compute these values step by step.

• $\binom{52}{8} = \frac{52!}{8!(52-8)!} = 75287520$
• $\binom{48}{8} = \frac{48!}{8!(48-8)!} = 75287520$
• $\binom{4}{1} = 4$
• $\binom{48}{7} = \frac{48!}{7!(48-7)!} = 12271512$

Now substitute these into the equation:

$\text{At least 2 queens} = 75287520 - \left(75287520 + 4 \times 12271512\right)$ $\text{At least 2 queens} = 75287520 - \left(75287520 + 49086048\right)$ $\text{At least 2 queens} = 75287520 - 124573568$ $\text{At least 2 queens} = 18520752$

Thus, the number of 8-card hands with at least 2 queens is 18,520,752.

Let me know if you need more clarification or if you'd like to explore similar problems!

Additional Questions:

1. How does the calculation change if you wanted a hand with at least 3 queens?
2. How would the number of ways to select a hand change if you were restricted to only face cards (Jacks, Queens, Kings)?
3. Can we use the same method to calculate hands with at least 2 cards of any specific rank (e.g., at least 2 Aces)?
4. How does the concept of combinations change when selecting from a deck of cards with different distributions (e.g., decks with jokers)?
5. What would happen if the deck size was increased from 52 to 60 cards?

Tip: When dealing with "at least" type problems, it's often easier to first calculate the total number of possibilities and then subtract the cases where the condition isn't met.