A standard pack of 52 cards consists of 4 suits, hearts, diamonds, clubs and spades. Each suit has 13 cards, from Ace to King. We deal randomly 5 cards from the deck of 52. 2 deals differing only by the order are considered the same.
How many different deals are there?
How many of them contain at most one ace?
How many contain cards of all four suits?
Do your latrobe homework yourself
To answer these questions, we'll need to use combinatorics, specifically counting principles such as combinations and permutations. Let's break it down step by step:
Question 1: How many different deals are there?
To determine the number of different deals, we'll use the concept of combinations. We want to choose 5 cards from a deck of 52 without regard to their order. The formula for combinations is nCr = n! / (r! * (n-r)!), where "n" is the total number of items, and "r" is the number we want to choose. In this case, we have 52 cards total, and we want to choose 5 cards, so the calculation becomes:
nCr = 52! / (5! * (52-5)!)
= 52! / (5! * 47!)
= (52 * 51 * 50 * 49 * 48) / (5 * 4 * 3 * 2 * 1)
= 2,598,960
Therefore, there are 2,598,960 different deals possible.
Question 2: How many of them contain at most one ace?
To count the number of deals that contain at most one ace, let's consider two cases: deals with no aces and deals with only one ace.
Case 1: No aces
Since there are 4 aces in the deck, we have 48 cards remaining to choose from (52 - 4). Using the combinations formula, we can calculate the number of deals with no aces:
nCr = 48! / (5! * (48-5)!)
= (48 * 47 * 46 * 45 * 44) / (5 * 4 * 3 * 2 * 1)
= 1,712,304
Case 2: One ace
Now, we have to consider the number of ways to choose 1 ace (4 options) and then 4 more cards from the remaining 48. Again, we can use the combinations formula:
nCr = 4! / (1! * (4-1)!) * 48! / (4! * (48-4)!)
= (4 * 48 * 47 * 46 * 45) / (3 * 2 * 1) * (48 * 47 * 46 * 45 * 44) / (4 * 3 * 2 * 1)
= 2,685,840
So, the total number of deals with at most one ace is the sum of the two cases:
Total = Case 1 + Case 2
= 1,712,304 + 2,685,840
= 4,398,144
Therefore, there are 4,398,144 deals that contain at most one ace.
Question 3: How many contain cards of all four suits?
To count the number of deals that contain cards of all four suits, consider that there is only one ace of each suit. We need to choose one ace from each suit (4 options) and then choose one more card from the remaining deck of 48. Using the combinations formula:
nCr = 4! / (4! * (4-4)!) * 48! / (1! * (48-1)!)
= (4 * 48) / (1) * (48 * 47 * 46 * 45 * 44) / (1 * 2 * 3 * 4 * 5)
= 15, 048
Therefore, there are 15,048 deals that contain cards of all four suits.