Problem 7
Question
Exer. 1-8: Find the number. $$ C(7,0) $$
Step-by-Step Solution
Verified Answer
The result of \( C(7, 0) \) is 1.
1Step 1: Understand the Combination Formula
The combination, denoted by \( C(n, r) \), represents the number of ways to choose \( r \) objects from \( n \) objects without regard to the order of selection. It is calculated using the formula: \[ C(n, r) = \frac{n!}{r!(n-r)!} \] where \(!\) represents factorial, which is the product of all positive integers up to that number.
2Step 2: Identify Values for \( n \) and \( r \)
In the problem given, \( C(7, 0) \), \( n \) is 7 and \( r \) is 0. This means we want to find the number of ways to select 0 objects from a set of 7 objects.
3Step 3: Calculate the Factorials
Calculate the factorials needed for the combination formula: \( 7! \), \( 0! \), and \( (7-0)! \). - \( 7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 5040 \) - \( 0! = 1 \) by definition of factorial - \( (7-0)! = 7! = 5040 \)
4Step 4: Substitute Values into the Formula
Substitute the known values into the combination formula: \[ C(7, 0) = \frac{7!}{0! \times 7!} \]
5Step 5: Simplify the Expression
Simplify the formula: \[ C(7, 0) = \frac{5040}{1 \times 5040} = \frac{5040}{5040} = 1 \] This shows there is 1 way to choose 0 objects from 7, which is to choose nothing.
Key Concepts
FactorialCombination FormulaPermutations
Factorial
Understanding factorial is essential for calculating combinations and permutations.
In simple terms, a factorial of a non-negative integer, denoted by \(!n!\), is the product of all positive integers less than or equal to \(n\). For instance, \(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\).
Some points to remember about factorials include:
Given the problem of finding \(C(7, 0)\), we calculated \(7!\) and \(0!\) as part of the steps in applying the combination formula. The concept of factorials is fundamental as it forms the basis for both permutations and combinations calculations.
In simple terms, a factorial of a non-negative integer, denoted by \(!n!\), is the product of all positive integers less than or equal to \(n\). For instance, \(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\).
Some points to remember about factorials include:
- Factorial of 0 is always 1, i.e., \(0! = 1\). This is a convention needed to ensure the consistency of mathematical formulas, especially in combination calculations.
- Factorials grow very quickly, making them extremely large, even for relatively small values of \(n\).
Given the problem of finding \(C(7, 0)\), we calculated \(7!\) and \(0!\) as part of the steps in applying the combination formula. The concept of factorials is fundamental as it forms the basis for both permutations and combinations calculations.
Combination Formula
A combination is about selecting items from a larger set, where the order of selection doesn't matter. This is captured in the formula: \[ C(n, r) = \frac{n!}{r!(n-r)!} \]
The elements here mean:
For example, with \(C(7, 0)\), you're finding the number of ways to choose 0 items from 7. The formula simplifies to:\[ C(7, 0) = \frac{7!}{0! \times 7!} \]Plugging in:\[ C(7, 0) = \frac{5040}{1 \times 5040} = 1 \]This calculation beautifully demonstrates the intuitive understanding that there's exactly 1 way to choose nothing from anything!
The elements here mean:
- \(n\) is the total number of items.
- \(r\) is the number of items to choose.
- \(!\) is the factorial symbol.
For example, with \(C(7, 0)\), you're finding the number of ways to choose 0 items from 7. The formula simplifies to:\[ C(7, 0) = \frac{7!}{0! \times 7!} \]Plugging in:\[ C(7, 0) = \frac{5040}{1 \times 5040} = 1 \]This calculation beautifully demonstrates the intuitive understanding that there's exactly 1 way to choose nothing from anything!
Permutations
Permutations and combinations are two sides of the same coin. While combinations focus on selection without regard for order, permutations do the opposite—they focus on ordering. \(P(n, r) = \frac{n!}{(n-r)!} \)
When calculating permutations, the order in which items are arranged matters, which is why the structure of the formula differs slightly from combinations.
For instance, if you were to find how many ways you could arrange 3 books from a total of 7, you’d use permutations. Simply put, permutations are for arranging, and combinations are for selecting. Understanding this difference helps you decide whether to use permutations or combinations when faced with real-world problems.
- \(n\) is the total number of items.
- \(r\) is the number of items to arrange.
When calculating permutations, the order in which items are arranged matters, which is why the structure of the formula differs slightly from combinations.
For instance, if you were to find how many ways you could arrange 3 books from a total of 7, you’d use permutations. Simply put, permutations are for arranging, and combinations are for selecting. Understanding this difference helps you decide whether to use permutations or combinations when faced with real-world problems.
Other exercises in this chapter
Problem 6
Exer. 3-10: Find the \(n\)th term, the fifth term, and the tenth term of the arithmetic sequence. $$ -6,-4.5,-3,-1.5, \ldots $$
View solution Problem 7
Find the \(n\)th term, the fifth term, and the eighth term of the geometric sequence. $$5,25,125,625, \ldots$$
View solution Problem 7
\(P(6,1)\)
View solution Problem 7
Exer. 1-26: Prove that the statement is true for every positive integer \(n\). $$ 1+2 \cdot 2+3 \cdot 2^{2}+\cdots+n \cdot 2^{n-1}=1+(n-1) \cdot 2^{n} $$
View solution