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:
(515)=51×50×49×48×475×4×3×2×1=2598960\binom{51}{5} = \frac{51 \times 50 \times 49 \times 48 \times 47}{5 \times 4 \times 3 \times 2 \times 1} = 2598960

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.

  1. Number of ways to choose a rank for the four cards:
    (131)=13\binom{13}{1} = 13
  2. Number of ways to choose 2 additional cards from the remaining 48 cards:
    (482)=48×472=1128\binom{48}{2} = \frac{48 \times 47}{2} = 1128 Thus, the total number of "Four of a Kind" hands is:
    13×1128=1466413 \times 1128 = 14664

(c) Exactly Two Queens

To have exactly two queens:

  1. Number of ways to select 2 queens from the 4 available:
    (42)=6\binom{4}{2} = 6
  2. Number of ways to choose 4 non-queen cards from the remaining 48 cards:
    (484)=48×47×46×454×3×2×1=194580\binom{48}{4} = \frac{48 \times 47 \times 46 \times 45}{4 \times 3 \times 2 \times 1} = 194580 Thus, the total number of hands with exactly two queens is:
    6×194580=11674806 \times 194580 = 1167480

(d) Contains No Hearts

To exclude hearts, choose all 6 cards from the remaining 39 non-heart cards:
(396)=39×38×37×36×35×346×5×4×3×2×1=575757\binom{39}{6} = \frac{39 \times 38 \times 37 \times 36 \times 35 \times 34}{6 \times 5 \times 4 \times 3 \times 2 \times 1} = 575757

Thus, 575,757 hands contain no hearts.


(e) Two Pairs

To form "Two Pairs":

  1. Choose 2 ranks for the pairs:
    (132)=13×122=78\binom{13}{2} = \frac{13 \times 12}{2} = 78
  2. Choose 2 cards (a pair) from each of the chosen ranks:
    (42)×(42)=6×6=36\binom{4}{2} \times \binom{4}{2} = 6 \times 6 = 36
  3. Choose 1 more card (not matching the ranks of the two pairs) from the remaining 44 cards:
    (441)=44\binom{44}{1} = 44 Thus, the total number of "Two Pairs" hands is:
    78×36×44=12355278 \times 36 \times 44 = 123552

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:

  1. How do the formulas for combinations and permutations differ in practical use?
  2. How does the distribution of suits impact poker hands?
  3. What is the probability of being dealt a "Four of a Kind" in poker?
  4. How are hands with a specific card calculated using complementary counting?
  5. 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