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?
(52!)/(5!47!) or 52C5 = 2,598,960 different ways a 5-card hand can be dealt from a deck of 52 cards.
(4C1)x(48C4) = the amount containing at most one ace.
(5C4)x(47C1) = amount containing cards of all four suits. (not very sure of this answer)
To find the answers to the questions, we need to use basic principles of counting and combinations.
1. How many different deals are there?
To calculate the total number of different deals, we need to evaluate how many ways we can choose 5 cards from a deck of 52 cards without regard to the order. This is a combination problem.
The number of different deals can be found using the formula for combinations, which is written as "nCk" or "n choose k." In this case, we want to calculate "52 choose 5", which can be written as C(52, 5).
The formula for a combination is: C(n, k) = n! / (k! * (n-k)!), where "!" represents the factorial.
Using this formula, we can calculate the number of different deals as follows:
C(52, 5) = 52! / (5! * (52-5)!) = (52 * 51 * 50 * 49 * 48) / (5 * 4 * 3 * 2 * 1) = 2,598,960
So, there are a total of 2,598,960 different deals.
2. How many of them contain at most one ace?
To calculate the number of deals containing at most one ace, we need to consider two cases:
Case 1: Deals with no aces:
In this case, we need to choose 5 cards from a deck of 48 cards (excluding the 4 aces). So, we use the combination formula C(48, 5).
C(48, 5) = 48! / (5! * (48-5)!) = (48 * 47 * 46 * 45 * 44) / (5 * 4 * 3 *2 * 1) = 1,712,304
Case 2: Deals with exactly one ace:
In this case, we need to choose one ace from the 4 available and four more cards from the remaining 48 cards (excluding the chosen ace). So, we use the combination formula C(4, 1) for choosing the ace and C(48, 4) for selecting the remaining cards.
C(4, 1) * C(48, 4) = (4 * 47 * 46 * 45 * 44) / (1 * 4 * 3 * 2 * 1) = 84,480
To find the total number of deals containing at most one ace, we sum the results from the two cases:
1,712,304 + 84,480 = 1,796,784
So, there are 1,796,784 different deals that contain at most one ace.
3. How many contain cards of all four suits?
To find the number of deals containing cards from all four suits, we need to evaluate the combinations for each suit separately.
For the first suit (hearts, for example), we have to choose one card from the 13 available in that suit. For the remaining suits, we need to choose one card from each.
C(13, 1) * C(13, 1) * C(13, 1) * C(13, 1) = 13 * 13 * 13 * 13 = 28,561
So, there are 28,561 different deals that contain cards from all four suits.