Math Problem Statement
Of 52 choose 6 = 20358520 different six-card poker hands, how many contains (a) The two of clubs? (b) Four of a kind? (c) Exactly two queens? (d) Contains no hearts? (e) Two pairs?
Solution
To solve each part, we will use combinatorics and logical reasoning.
(a) Contains the Two of Clubs
If the two of clubs is included, we need to choose 5 more cards from the remaining 51 cards. The total combinations for the remaining cards are:
So, 2,598,960 hands contain the two of clubs.
(b) Four of a Kind
To form "Four of a Kind," choose one rank for the four cards and then choose two other cards from the remaining 48 cards.
- Number of ways to choose a rank for the four cards:
- Number of ways to choose 2 additional cards from the remaining 48 cards:
Thus, the total number of "Four of a Kind" hands is:
(c) Exactly Two Queens
To have exactly two queens:
- Number of ways to select 2 queens from the 4 available:
- Number of ways to choose 4 non-queen cards from the remaining 48 cards:
Thus, the total number of hands with exactly two queens is:
(d) Contains No Hearts
To exclude hearts, choose all 6 cards from the remaining 39 non-heart cards:
Thus, 575,757 hands contain no hearts.
(e) Two Pairs
To form "Two Pairs":
- Choose 2 ranks for the pairs:
- Choose 2 cards (a pair) from each of the chosen ranks:
- Choose 1 more card (not matching the ranks of the two pairs) from the remaining 44 cards:
Thus, the total number of "Two Pairs" hands is:
Final Answers:
(a) 2,598,960
(b) 14,664
(c) 1,167,480
(d) 575,757
(e) 123,552
Would you like detailed steps for any part? Here are 5 questions for exploration:
- How do the formulas for combinations and permutations differ in practical use?
- How does the distribution of suits impact poker hands?
- What is the probability of being dealt a "Four of a Kind" in poker?
- How are hands with a specific card calculated using complementary counting?
- How does adding constraints to card selection affect combinatorial outcomes?
Tip: When solving combinatorics problems, start by identifying constraints and breaking the problem into smaller, manageable parts.
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Math Problem Analysis
Mathematical Concepts
Combinatorics
Probability
Card combinations
Formulas
Combination formula: C(n, k) = n! / (k! * (n-k)!)
Theorems
Basic principles of combinatorics
Suitable Grade Level
Grades 10-12
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