In a standard deck of cards there are four eights and four sevens.\nHow many different 5 card hands can be dealt that consist of exactly three eights and two sevens?

Introduction / Context:
This question is about counting specific types of card hands using combinations. We are asked to find how many 5 card hands contain exactly three cards of one rank (eight) and exactly two cards of another rank (seven). The key is to recognise that within each rank we are choosing a subset of the available cards.

Given Data / Assumptions:

• Standard deck: 52 playing cards.
• Number of eights in the deck = 4.
• Number of sevens in the deck = 4.
• We must form a hand of exactly 5 cards.
• Hand must contain exactly three eights and exactly two sevens.
• Order of cards in a hand does not matter.

Concept / Approach:
To construct such a hand, we must independently choose which eights appear and which sevens appear. Since there are 4 cards of each rank, the number of ways to choose 3 eights is C(4,3), and the number of ways to choose 2 sevens is C(4,2). Because these choices are independent, we multiply the two combination counts to get the total number of distinct hands fitting the requirement.

Step-by-Step Solution:
Number of ways to choose 3 eights from 4 eights = C(4,3). Compute C(4,3) = 4! / (3! * 1!) = 4. Number of ways to choose 2 sevens from 4 sevens = C(4,2). Compute C(4,2) = 4! / (2! * 2!) = 6. Total hands with exactly three eights and two sevens = C(4,3) * C(4,2) = 4 * 6. Compute 4 * 6 = 24.

Verification / Alternative check:
Since a hand must contain exactly 3 + 2 = 5 cards, and there are only two ranks involved, there is no flexibility to include other ranks. Each valid hand is therefore uniquely determined by which specific three eights and which specific two sevens it contains. There is no risk of double counting because we never change the hand structure, only choose specific cards within each group. The multiplication rule thus gives the correct answer. The small final number reflects the strict requirement.

Why Other Options Are Wrong:

• 36: This might come from incorrectly using C(4,3) * C(4,3) or another wrong combination count.
• 72: This is three times the correct answer, perhaps from mistakenly multiplying by an extra factor such as 3!.
• 16: This is too small and suggests simply 4 * 4, which would be wrong because we must choose combinations, not count ordered pairs.

Common Pitfalls:
Typical mistakes include treating eights and sevens as if any number of each can appear, which leads to unwanted hands with different compositions, or using permutations instead of combinations. Some students also forget the condition "exactly three" eights and "exactly two" sevens and accidentally include cases with four eights and one seven or similar variations. Clearly restating the required counts before computing combinations helps avoid these errors.

Final Answer:
The number of different 5 card hands that contain exactly three eights and two sevens is 24.