Problem 22
Question
A standard deck contains 52 different cards. In how many ways can you select 5 cards from the deck?
Step-by-Step Solution
Verified Answer
There are 2,598,960 ways to select 5 cards from a deck of 52 cards.
1Step 1: Understand the Problem
To solve this problem, we need to find the number of ways to select 5 cards from a deck of 52 cards. This scenario represents a combination where the order of selection does not matter.
2Step 2: Apply Combination Formula
The formula for combinations is \( C(n, r) = \frac{n!}{r!(n-r)!} \) where \( n \) is the total number of items to choose from, and \( r \) is the number of items to choose. In this problem, \( n = 52 \) and \( r = 5 \).
3Step 3: Calculate the Factorials
Compute the factorials needed for the formula: \( 52! \), \( 5! \), and \( (52-5)! = 47! \). However, for practical calculation, manual calculation of full \( 52! \) is not needed as it will be reduced in the combination formula.
4Step 4: Plug Values into the Combination Formula
Substitute the values into the combination formula: \[ C(52, 5) = \frac{52!}{5! \times 47!} \]
5Step 5: Simplify the Formula
Recognize that \( 52! = 52 \times 51 \times 50 \times 49 \times 48 \times 47! \), allowing \( 47! \) to cancel out in the numerator and denominator:\[ C(52, 5) = \frac{52 \times 51 \times 50 \times 49 \times 48}{5 \times 4 \times 3 \times 2 \times 1} \]
6Step 6: Calculate the Final Result
Perform the arithmetic:- \( 52 \times 51 \times 50 \times 49 \times 48 = 311875200 \)- \( 5 \times 4 \times 3 \times 2 \times 1 = 120 \)- Divide them: \( \frac{311875200}{120} = 2598960 \)Therefore, the number of ways to select 5 cards from a deck of 52 is 2,598,960.
Key Concepts
Combination FormulaFactorialsBasic Probability
Combination Formula
Combinations are essential when it comes to probabilities because they help us determine how many ways we can choose a subset of items from a larger set, where the order of items does not matter. This is often represented by the formula \[ C(n, r) = \frac{n!}{r!(n-r)!} \] Here, **\( n \)** is the total number of items available, and **\( r \)** is the number of items we wish to choose.
- Key Point: The combination formula calculates the number of different groups or sets that can be formed from a larger set.
- Order Doesn’t Matter: If the order was important, we would be dealing with permutations instead of combinations.
- Real Life Usage: Anytime you need to select members for a team, choose cards, or form committees where order isn't relevant, combinations are your tool.
Factorials
Factorials are a vital mathematical operation in combinatorics and probability, often represented by an exclamation mark '!', such as \( n! \). It means multiplying together all whole numbers from 1 up to \( n \).
- Definition: \( n! = n \times (n-1) \times (n-2) \times ... \times 2 \times 1 \)
- The Simple Case: 0! is defined to be 1. This might seem odd, but it's essential for mathematical consistency.
- Usage: Factorials are used to find out how many different ways items can be arranged, and are crucial in the combination and permutation formulas.
Basic Probability
Basic probability helps us understand the likelihood of events occurring. Probability follows a simple rule: it's the ratio of the number of favorable outcomes to the total number of possible outcomes.
- Formula: Probability (\( P \)) = \( \frac{\text{Number of Favorable Outcomes}}{\text{Total Number of Possible Outcomes}} \)
- Between 0 and 1: Probability values range between 0 and 1, where 0 means the event is impossible, and 1 means it's certain.
- Complementary Rule: The probability that an event does not happen is \( 1 - P \).
Other exercises in this chapter
Problem 22
Toss three fair coins and find the probability of no heads.
View solution Problem 22
You pick 3 cards from a standard deck of 52 cards. Find the probability that the third card is an ace. Compare this with the probability that the first card is
View solution Problem 22
The following data represent a sample from a normal distribution with mean 0 and variance \(1:\) $$ \begin{array}{l} -1.18,0.52,0.36,-0.16,0.92 \\ 0.68,-0.61,-0
View solution Problem 22
Suppose a genotypic trait is controlled by 90 loci. Each locus, independently of all others, contributes to the genotypic value of the trait either \(1.1\) with
View solution