Combinatorial Problems in a Standard 52-Card Deck: Analyzing Different Suit Configurations

This article explores various combinatorial problems related to selecting cards from a standard 52-card deck. The focus will be on calculating the number of different choices possible under specific conditions. These include scenarios where all 7 cards are of the same suit, at least 5 cards are of the same suit, and where 4 cards can be of the same suit while the other 3 are from different suits. The methods used to solve these problems involve combinatorics, specifically combinations, which are denoted as Cnk or binom{n}{k}, where n is the total number of items to choose from, and k is the number of items to choose.

Problem A: All 7 Cards Must Be of the Same Suit

Step 1: Choosing a Suit There are 4 suits (hearts, diamonds, clubs, spades) from which we can choose. Therefore, there are 4 ways to choose a suit.

Step 2: Choosing 7 Cards from That Suit Each suit contains 13 cards. The number of ways to choose 7 cards from 13 is given by the combination formula:

[ C_{13}^{7} frac{13!}{7!(13-7)!} frac{13!}{7!6!} 1716 ]

Final Calculation Multiplying the number of ways to choose a suit by the number of ways to choose 7 cards from that suit, we get:

[ 4 times 1716 6864 ] Therefore, the total number of ways to choose 7 cards of the same suit is 6864.

Problem B: At Least 5 Cards Must Be of the Same Suit

This problem can be broken down into three cases: exactly 5, 6, and 7 cards of the same suit.

Case 1: Exactly 5 Cards of One Suit, 2 Cards From Other Suits

Choosing the Suit for 5 Cards There are 4 ways to choose the suit for the 5 cards. Choosing 5 Cards from That Suit The number of ways to choose 5 cards from the 13 available is given by:

[ C_{13}^{5} frac{13!}{5!(13-5)!} frac{13!}{5!8!} 1287 ]

Choosing 2 Suits for the Other 2 Cards Choose 2 suits from the remaining 3 in:

[ C_{3}^{2} frac{3!}{2!(3-2)!} 3 ]

Choosing 1 Card from Each of the 2 Chosen Suits Each suit has 13 cards, so the number of ways to choose 1 card from each of the 2 suits is:

[ 13 times 13 169 ]

Total for Case 1 Multiplying all these values together:

[ 4 times 1287 times 3 times 169 2607828 ]

Case 2: Exactly 6 Cards of One Suit, 1 Card From Another Suit

Choosing the Suit for 6 Cards There are 4 ways to choose the suit for the 6 cards. Choosing 6 Cards from That Suit The number of ways to choose 6 cards from the 13 available is given by:

[ C_{13}^{6} frac{13!}{6!(13-6)!} frac{13!}{6!7!} 1716 ]

Choosing 1 Card from One of the Remaining 3 Suits There are 3 ways to choose the suit for the remaining card and 13 ways to choose the card from that suit, so the total is:

[ 3 times 13 39 ]

Total for Case 2 Multiplying all these values together:

[ 4 times 1716 times 39 267744 ]

Case 3: All 7 Cards of One Suit

Combinations for Case 3 This was calculated in Problem A, giving 6864.

Total for Part B Summing up all the cases, we get:

[ 2607828 267744 6864 2883436 ]

Problem C: 4 Cards Can Be of the Same Suit, Other 3 of Different Suits

Step 1: Choosing the Suit for the 4 Cards There are 4 ways to choose the suit for the 4 cards.

Step 2: Choosing 4 Cards from That Suit The number of ways to choose 4 cards from the 13 available is given by:

[ C_{13}^{4} frac{13!}{4!(13-4)!} frac{13!}{4!9!} 715 ]

Step 3: Choosing 3 Different Suits from the Remaining 3 Suits Since we need all 3 remaining suits, there is only 1 way to choose the remaining suits.

Step 4: Choosing 1 Card from Each of the 3 Chosen Suits Each suit has 13 cards, so the number of ways to choose 1 card from each of the 3 suits is:

[ 13 times 13 times 13 2197 ]

Total for Part C Multiplying all these values together:

[ 4 times 715 times 2197 6285620 ]

Final Answers

Problem A: 6864

Problem B: 2883436

Problem C: 6285620