From a standard deck of 52 cards, a 5-card hand is dealt. If the queen of spades and the four of diamonds must both be included in the hand, how many distinct 5-card hands are possible?

Introduction / Context:
This card problem asks you to count how many 5 card hands contain two specific cards: the queen of spades and the four of diamonds. Such problems highlight the idea of fixing certain required cards and then choosing the remaining cards freely from the rest of the deck. This is a very common pattern in conditional counting problems in probability and combinatorics.

Given Data / Assumptions:

• Standard deck with 52 distinct cards.
• We deal a 5 card hand.
• Two specific cards, the queen of spades and the four of diamonds, must be in every valid hand.
• The remaining 3 cards can be any cards except those two already required.
• Order of cards in the hand does not matter; only the set of 5 cards matters.

Concept / Approach:
We first fix the two required cards in the hand. Once these are chosen, they occupy 2 of the 5 positions. We then need to choose the remaining 3 cards from the deck excluding these two cards. That leaves 52 - 2 = 50 cards available. The number of ways to choose 3 cards from these 50, with order not mattering, is given by the combination 50C3.

Step-by-Step Solution:
Step 1: Include the queen of spades in the hand. Step 2: Include the four of diamonds in the hand. Step 3: Now 2 of the 5 cards are fixed; we still need 3 more cards. Step 4: Cards remaining in the deck that are eligible: 52 - 2 = 50. Step 5: The number of ways to choose 3 more cards from these 50 is 50C3. Step 6: So the number of distinct 5 card hands containing both fixed cards is 50C3.

Verification / Alternative check:
If we want a numerical value, compute 50C3: 50! / (3! * 47!) = (50 * 49 * 48) / (3 * 2 * 1) = 19600. Any hand counted this way has the two required cards plus 3 chosen from the remaining 50, and no hand is counted twice. The combination expression 50C3 exactly represents this process, so it is the correct form among the options provided.

Why Other Options Are Wrong:
52C5: Counts all possible 5 card hands without any requirement, including many that do not contain both required cards. 52C4: Would correspond to choosing 4 cards from 52, which does not match the structure of our 5 card hand with two fixed cards. 50C4: This would arise if we were choosing 4 remaining cards instead of 3, but that would create a 6 card hand together with the 2 fixed cards. Only 50C3 correctly models choosing the remaining 3 cards from the 50 not fixed cards.

Common Pitfalls:
Some learners forget to subtract the two fixed cards from the deck and mistakenly use 52C3. Others try to count the ways to place the fixed cards in positions inside the hand, but because a hand is just a set of cards, that positional counting is unnecessary and leads to overcounting. The correct strategy is always to fix required elements and then choose remaining elements from what is left.

Final Answer:
The number of distinct 5 card hands that contain both the queen of spades and the four of diamonds is 50C3 (numerically equal to 19600).