Problem 49
Question
These problems involve combinations. Draw Poker Hands How many different five- card hands can be dealt from a deck of 52 cards?
Step-by-Step Solution
Verified Answer
There are 2,598,960 different five-card hands possible from a deck of 52 cards.
1Step 1: Understand the problem
We need to find the number of different ways to select 5 cards from a deck of 52 cards. This is a problem of combinations because the order in which the cards are drawn does not matter.
2Step 2: Set up the combination formula
To find the number of combinations, we use the formula for combinations: \( C(n, r) = \frac{n!}{r!(n-r)!} \), where \( n \) is the total number of items, and \( r \) is the number of items to choose.
3Step 3: Identify the given values
In this problem, \( n = 52 \) because there are 52 cards in a deck, and \( r = 5 \) because we need to choose 5 cards.
4Step 4: Plug values into the combination formula
Substitute \( n = 52 \) and \( r = 5 \) into the combination formula: \( C(52, 5) = \frac{52!}{5!(52-5)!} \).
5Step 5: Calculate the factorial values
Calculating the factorials needed: \( 52! \), \( 5! \), and \( 47! \). Note that \( 52! \) can be simplified using \( (52 \times 51 \times 50 \times 49 \times 48) \) because \( 47! \) cancels out of the numerator and denominator.
6Step 6: Simplify the expression
Simplify \( \frac{52 \times 51 \times 50 \times 49 \times 48}{5 \times 4 \times 3 \times 2 \times 1} \). This equals \( \frac{311875200}{120} \).
7Step 7: Perform the final division
Divide \( 311875200 \) by \( 120 \) to get the final result: \( 2598960 \).
8Step 8: Conclusion
There are 2,598,960 different five-card hands that can be dealt from a deck of 52 cards.
Key Concepts
Combination FormulaFactorials in MathematicsCombinatorial Problems
Combination Formula
The combination formula is a fundamental concept in combinatorics, which allows us to calculate the number of ways to choose a subset of items from a larger set. When the order of selection does not matter, we use combinations instead of permutations.
The formula to determine the number of combinations, symbolized as \( C(n, r) \), is given by:
When solving problems with combinations, it is crucial to establish whether order matters. For instance, in card games, the order cards are dealt usually does not matter, so combinations are used.
The formula to determine the number of combinations, symbolized as \( C(n, r) \), is given by:
- \( C(n, r) = \frac{n!}{r!(n-r)!} \)
- \(n\) is the total number of items to pick from.
- \(r\) is the number of items to be chosen.
When solving problems with combinations, it is crucial to establish whether order matters. For instance, in card games, the order cards are dealt usually does not matter, so combinations are used.
Factorials in Mathematics
Factorials, denoted by \(!\), represent the product of an integer and all positive integers below it. For example:
The concept of factorials is crucial for simplifying expressions when computing combinations because many terms can cancel out between the numerator and denominator. In our five-card hands example, the factorial calculations for \(52!\), \(5!\), and \((52-5)!\) are needed. Simplifying \(\frac{52!}{47!}\) involves calculating only \(52 \times 51 \times 50 \times 49 \times 48\), making the computation more manageable.
Factorials grow surprisingly fast; for instance, \(10! = 3,628,800\). Understanding how to work with and simplify factorials is essential in many combinatorial problems.
- \(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\)
The concept of factorials is crucial for simplifying expressions when computing combinations because many terms can cancel out between the numerator and denominator. In our five-card hands example, the factorial calculations for \(52!\), \(5!\), and \((52-5)!\) are needed. Simplifying \(\frac{52!}{47!}\) involves calculating only \(52 \times 51 \times 50 \times 49 \times 48\), making the computation more manageable.
Factorials grow surprisingly fast; for instance, \(10! = 3,628,800\). Understanding how to work with and simplify factorials is essential in many combinatorial problems.
Combinatorial Problems
Combinatorial problems focus on counting, arranging, and optimizing objects within a set. Understanding how combinations and permutations work is vital for solving these types of problems.
By using combinations, we solve problems where the order of selection does not matter. Conversely, permutations are used when the order does matter. Combinatorial problems are omnipresent in fields including mathematics, computer science, and operations research, demonstrating how items can be arranged or grouped in distinct ways.
Real-life applications involve situations like forming committees, scheduling, or organizing competitions. Let's consider our original exercise: determining five-card hands from a deck of cards. Here, we look into possible groupings without attention to the order, hence applying the combination formula.
Focusing on combinations, alongside efficient factorial handling, such as partial factorials, is important in managing complex calculations efficiently within combinatorial contexts.
By using combinations, we solve problems where the order of selection does not matter. Conversely, permutations are used when the order does matter. Combinatorial problems are omnipresent in fields including mathematics, computer science, and operations research, demonstrating how items can be arranged or grouped in distinct ways.
Real-life applications involve situations like forming committees, scheduling, or organizing competitions. Let's consider our original exercise: determining five-card hands from a deck of cards. Here, we look into possible groupings without attention to the order, hence applying the combination formula.
Focusing on combinations, alongside efficient factorial handling, such as partial factorials, is important in managing complex calculations efficiently within combinatorial contexts.
Other exercises in this chapter
Problem 47
These problems involve combinations. Committee In how many ways can a committee of three members be chosen from a club of 25 members?
View solution Problem 48
These problems involve combinations. Choosing a Group In how many ways can six people be chosen from a group of ten?
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Roulette An American roulette wheel has 38 slots. Two of the slots are numbered 0 and \(00,\) and the rest are numbered from 1 to \(36 .\) Find the probability
View solution Problem 50
These problems involve combinations. Stud Poker Hands How many different seven-card hands can be picked from a deck of 52 cards?
View solution