To solve this problem, we need to determine how many 7-card hands consist of exactly 2 kings and 3 queens from a standard deck of 52 cards.

Step 1: Choose the Kings

In a standard deck, there are 4 kings (one from each suit: clubs, diamonds, hearts, and spades). To select exactly 2 kings from these 4, we use the combination formula:

$\binom{4}{2} = \frac{4!}{2!(4-2)!} = \frac{4 \times 3}{2 \times 1} = 6$

So, there are 6 ways to choose 2 kings.

Step 2: Choose the Queens

Similarly, there are 4 queens in the deck, and we need to select exactly 3 queens. Using the combination formula again:

$\binom{4}{3} = \frac{4!}{3!(4-3)!} = \frac{4 \times 3 \times 2}{3 \times 2 \times 1} = 4$

There are 4 ways to choose 3 queens.

Step 3: Choose the Remaining 2 Cards

We've already selected 5 cards (2 kings and 3 queens). Now, we need to choose 2 more cards from the remaining 44 cards in the deck (since there are 52 cards in total, and we have already excluded the 4 kings and 4 queens, leaving 44 other cards).

The number of ways to choose 2 cards from these remaining 44 cards is:

$\binom{44}{2} = \frac{44 \times 43}{2 \times 1} = 946$

Step 4: Multiply the Results

To find the total number of 7-card hands that consist of exactly 2 kings and 3 queens, we multiply the number of ways to choose the kings, the number of ways to choose the queens, and the number of ways to choose the remaining 2 cards:

$6 \times 4 \times 946 = 22,704$

Final Answer:

There are 22,704 different 7-card hands that consist of exactly 2 kings and 3 queens.

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Would you like more details or have any questions?

Here are 5 related questions to consider:

1. How many 7-card hands consist of exactly 2 aces and 3 jacks?
2. What is the probability of drawing a 7-card hand with exactly 2 kings and 3 queens?
3. How many 5-card hands consist of exactly 1 king and 2 queens?
4. What is the total number of possible 7-card hands from a deck of 52 cards?
5. How many 7-card hands consist of exactly 3 aces and 2 kings?

Tip: Combinations are used when the order of selection doesn't matter; if the order matters, permutations are used instead!