To find the number of different hands possible where a hand of 8 cards contains exactly 3 sevens from a regular deck of 52 cards, we can approach the problem step by step:

Step 1: Choose 3 sevens.

• A standard deck has four sevens (7♠, 7♥, 7♦, 7♣).
• We need to select 3 of these four sevens.

The number of ways to choose 3 sevens from 4 is given by the combination:

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

Step 2: Choose 5 non-sevens.

• After choosing 3 sevens, there are $52 - 4 = 48$ non-seven cards left in the deck.
• We need to choose 5 cards from these remaining 48 cards.

The number of ways to choose 5 cards from 48 is given by the combination:

$\binom{48}{5} = \frac{48!}{5!(48-5)!}$

Step 3: Multiply the results.

• The total number of different hands is the product of the two combinations:

$\text{Total hands} = \binom{4}{3} \times \binom{48}{5}$

Substituting the values:

$\text{Total hands} = 4 \times \binom{48}{5} = 4 \times 1,712,304 = 6,849,216$

Final Answer:

There are 6,849,216 different hands possible with exactly 3 sevens in an 8-card hand.

Would you like more details on combinations or any step here?

Related questions:

1. How many hands are possible if the hand contains exactly 2 sevens?
2. What is the probability of getting exactly 3 sevens in an 8-card hand?
3. How many different hands are possible if the hand contains at least one seven?
4. How many different hands are possible if the hand contains exactly 4 sevens?
5. What is the probability of getting no sevens in an 8-card hand?

Tip: The combination formula $\binom{n}{k}$ is useful for counting selections without regard to order.