The problem asks for the number of different ways to deal a hand of 5 cards from a standard deck of 52 cards, where the order does not matter. This is a combination problem because we are choosing 5 cards out of 52 without regard to the order in which they are dealt.

The formula for combinations is:

$C(n, k) = \frac{n!}{k!(n-k)!}$

Where:

• $n$ is the total number of items (in this case, 52 cards),
• $k$ is the number of items to choose (in this case, 5 cards).

Substitute the values into the formula:

$C(52, 5) = \frac{52!}{5!(52-5)!} = \frac{52!}{5!47!}$

Now, simplifying without calculating the full factorial:

$C(52, 5) = \frac{52 \times 51 \times 50 \times 49 \times 48}{5 \times 4 \times 3 \times 2 \times 1} = \frac{311,875,200}{120} = 2,598,960$

Thus, there are 2,598,960 different ways to deal a hand of 5 cards from a standard deck of 52 cards if the order is not considered.

Would you like further details or have any other questions? Here are five related questions to think about:

1. What if the order of the cards did matter? How would you calculate that?
2. How many ways are there to deal 2 hands of 5 cards each?
3. How many ways can you deal a hand with exactly 2 aces?
4. How does the probability of getting a flush (all cards of the same suit) affect this scenario?
5. How many different 5-card poker hands exist based on typical poker rules?

Tip: When solving counting problems, remember the difference between combinations (order doesn't matter) and permutations (order does matter).