Problem 63
Question
From a group of \( 40 \) people, a jury of \( 12 \) people is to be selected. In how many different ways can the jury be selected?
Step-by-Step Solution
Verified Answer
The jury can be selected in 5,586,853,480 different ways.
1Step 1: Identify the Known Variables
In this problem, 'n' is the total number of people which is 40, and 'r' is the number of people to be selected in the jury, which is 12.
2Step 2: Apply the Combination Formula
We will use the combination formula \( C(n, r) = \frac{n!}{r!(n-r)!} \) and plug in our values of 'n' and 'r'. The solution then becomes \( C(40, 12) = \frac{40!}{12!(40-12)!} \)
3Step 3: Simplify the Factorials
Calculate the factorial of each number in the formula.
4Step 4: Calculate Combination
Perform the division to find the solution.
Key Concepts
Combination FormulaFactorialPermutation
Combination Formula
Choosing a subset of items from a larger set without concern for the order in which they're selected is known as a combination. This is where the Combination Formula comes into play.
This yields the total number of possible ways to select 12 people from 40. The beauty of the combination formula is that it focuses solely on selection, disregarding the sequence in which items are picked.
- The formula is written as \( C(n, r) = \frac{n!}{r!(n-r)!} \), where \( n \) represents the total number of items to choose from, and \( r \) stands for the number of items to select.
- Using the combination formula is perfect for scenarios like choosing members of a jury, a team, or selecting toppings for a pizza where order does not matter.
This yields the total number of possible ways to select 12 people from 40. The beauty of the combination formula is that it focuses solely on selection, disregarding the sequence in which items are picked.
Factorial
Factorial is a fundamental concept in combinatorics, which involves multiplying a series of descending natural numbers. It is expressed as \( n! \) (read as 'n factorial'). The definition states:
For example, calculating 12! would involve multiplying \( 12 \times 11 \times 10 \times ... \times 1 \).
In the context of our combination problem, we used factorials to compute the number of different ways to choose people from a group by applying them within the formula \( \frac{40!}{12! \times 28!} \). Factorials can be understood more easily when breaking them down into their components, ensuring calculations are done step-by-step.
- For any positive integer \( n \), \( n! = n \times (n−1) \times (n−2) \times \, ... \, \times 1 \).
- The factorial of 0 is defined as 1, which helps in certain calculations and ensures consistency within equations.
For example, calculating 12! would involve multiplying \( 12 \times 11 \times 10 \times ... \times 1 \).
In the context of our combination problem, we used factorials to compute the number of different ways to choose people from a group by applying them within the formula \( \frac{40!}{12! \times 28!} \). Factorials can be understood more easily when breaking them down into their components, ensuring calculations are done step-by-step.
Permutation
Permutation is another essential concept in combinatorics, used when the order of selection matters. Unlike combinations, where order doesn't compare to permutations,
In contrast to the jury selection problem, if the order of chosen people mattered, permutations would be the tool to use. However, for non-ordered selections like the jury example, combinations take precedence. Understanding when to use permutations versus combinations relies partially on whether the order is significant in the given scenario.
- Permutation counts different ways to arrange a subset of items within a larger set.
- The permutation formula is \( P(n, r) = \frac{n!}{(n-r)!} \), where order plays a crucial role in determining different arrangements.
In contrast to the jury selection problem, if the order of chosen people mattered, permutations would be the tool to use. However, for non-ordered selections like the jury example, combinations take precedence. Understanding when to use permutations versus combinations relies partially on whether the order is significant in the given scenario.
Other exercises in this chapter
Problem 62
In Exercises 59 - 66, find the indicated \( n \)th partial sum of the arithmetic sequence. \( 0.5, 1.3, 2.1, 2.9, \cdots , n = 10 \)
View solution Problem 62
In Exercises 47-62, write an expression for the apparent \(n\)th term of the sequence. (Assume that \( n \) begins with 1.) \( 1 + \dfrac{1}{2}, 1 + \dfrac{3}{4
View solution Problem 63
In Exercises 61 - 66, use the Binomial Theorem to expand and simplify the expression. \( \left(x^{2/3} - y^{1/3}\right)^3 \)
View solution Problem 63
In Exercises 59 - 64, decide whether the sequence can be represented perfectly by a linear or a quadratic model. If so,find the model. \( -2, 1, 6, 13, 22, 33,
View solution