In a standard deck of cards, how many distinct 5-card hands form a straight using the ranks 4, 5, 6, 7, and 8 (that is, starting with 4 and ending with 8), if suits can be chosen freely and the order of cards in the hand does not matter?

Introduction / Context:
This question is about counting specific poker type hands: 5-card straights that run from 4 to 8. The focus is on how many distinct hands can be made with those exact ranks while suits can vary freely. This is a typical intermediate card combinatorics question where ranks are fixed but suits are flexible. Understanding such counting is helpful for probability problems involving straights and related events in card games.

Given Data / Assumptions:

• We use a standard 52-card deck with 4 suits.
• The straight must use the ranks 4, 5, 6, 7, and 8.
• For each rank, any of the four suits (clubs, diamonds, hearts, spades) is allowed.
• The order of cards in the hand does not matter; hands are just sets of cards.
• No card can repeat, so each chosen card has a distinct combination of rank and suit.

Concept / Approach:
The ranks are fixed: we must take exactly one card of rank 4, one of rank 5, one of rank 6, one of rank 7, and one of rank 8. For each rank, there are 4 possible suits. The choices for different ranks are independent of each other. Therefore, the total number of distinct hands is the product of suit choices across all five ranks.

Step-by-Step Solution:
Step 1: For rank 4, there are 4 possible cards (one in each suit). Step 2: For rank 5, there are again 4 possible cards. Step 3: Similarly, ranks 6, 7, and 8 each contribute 4 possible cards. Step 4: Because we choose exactly one card of each rank, multiply the choices: 4 * 4 * 4 * 4 * 4. Step 5: Compute 4^5 = 4 * 4 * 4 * 4 * 4 = 1024. Step 6: The order of cards does not matter, but each 5-card collection here is already a unique hand because the composition (specific rank and suit of each card) is different. Step 7: Thus, total number of such straight hands is 1024.

Verification / Alternative check:
We can think of this as a function assignment: for each of the 5 distinct ranks, choose one of 4 suits, which gives 4^5 possibilities. There is no overlap, since each resulting set of cards is unique. No division by 5! is required because we are not counting sequences; we are counting distinct sets of cards already specified by their ranks. This reasoning confirms 1024 as the correct answer.

Why Other Options Are Wrong:
1296: Equal to 6^4, which does not relate correctly to the structure of this problem. 1094: A random looking number, not a power or simple product relevant to the scenario. 1200: Does not arise from any valid combination or power count for five ranks with four suits. None of these are equal to 4^5, which is the correct computation.

Common Pitfalls:
A typical mistake is to treat this as 52C5, which would count all possible 5-card hands instead of only straights from 4 to 8. Another error is to think that one must divide by 5! to account for order, forgetting that by directly counting suit assignments per rank we already avoid overcounting again. Some learners also confuse this with counting the number of different rank sequences, but here the rank pattern is completely fixed, so we only vary suits.

Final Answer:
The number of distinct 5-card hands that form a straight using ranks 4, 5, 6, 7, and 8 is 1024.