CK_OI diamonds (red), spades (black) and clubs (black). a) How many cards must be selected from a standard deck of cards to guarantee that at least 3 cards of the same suit are selected? Explain your answer. How many ways are there to select a pair of cards from a standard deck of cards such that one of the cards is red and the other one is black? Your answer can contain factorial or power expressions. Explain your answer. How many ways are there to divide a standard deck of cards over 4 players? Your answer can contain factorial or power expressions. Explain your answer. None
The Correct Answer and Explanation is:
Let’s tackle each part of the question one by one, with clear answers and detailed explanations.
a) How many cards must be selected from a standard deck to guarantee at least 3 cards of the same suit are selected?
Answer: 9 cards
Explanation:
A standard deck has 4 suits: hearts ♥️, diamonds ♦️ (both red), and clubs ♣️, spades ♠️ (both black). Each suit has 13 cards.
To guarantee that at least 3 cards of the same suit are selected, we apply the Pigeonhole Principle. This principle tells us how items (cards) can be distributed into containers (suits) before a particular condition must occur.
Worst-case scenario:
- 2 cards of each suit (2 hearts, 2 diamonds, 2 spades, 2 clubs) = 8 cards
- No suit yet has 3 cards.
Now, on drawing the 9th card, at least one suit will have 3 cards, no matter which suit it is.
✅ Therefore, 9 cards are needed to guarantee 3 of the same suit.
b) How many ways are there to select a pair of cards from a standard deck such that one is red and one is black?
Answer: 26 × 26 = 676 ways
Explanation:
There are:
- 26 red cards (13 hearts + 13 diamonds)
- 26 black cards (13 spades + 13 clubs)
To form a pair with one red and one black card:
- Choose 1 red card in 26 ways
- Choose 1 black card in 26 ways
Total ways = 26 × 26 = 676
Alternatively, expressed in factorials: (261)×(261)=26×26\binom{26}{1} \times \binom{26}{1} = 26 \times 26
✅ Hence, there are 676 distinct red-black card pairs.
c) How many ways are there to divide a standard deck of cards over 4 players?
Answer: 52!(13!)4\frac{52!}{(13!)^4}
Explanation:
We are dividing 52 cards into 4 hands of 13 cards each, with each hand going to one player.
- First, count all ways to arrange 52 cards: 52!
- But within each player’s hand, the order of cards doesn’t matter: divide by 13! for each hand.
- Also, since players are distinct (say Player A, B, C, D), the division is unique per player.
So the number of ways is: 52!13!×13!×13!×13!=52!(13!)4\frac{52!}{13! \times 13! \times 13! \times 13!} = \frac{52!}{(13!)^4}
✅ Thus, there are 52!(13!)4\frac{52!}{(13!)^4} ways to deal 52 cards evenly to 4 players.