Problem 42
Question
How many arrangements are there of 6 different books on a shelf?
Step-by-Step Solution
Verified Answer
There are 720 arrangements of the 6 books.
1Step 1: Understanding the Problem
We need to find out how many different ways we can arrange 6 distinct books on a shelf.
2Step 2: Recognize the Permutation
When dealing with arrangements, especially where order matters, we're typically dealing with permutations. In this scenario, we are arranging 6 distinct items (books).
3Step 3: Apply Permutation Formula
To find the number of permutations of 6 items, we use the formula for permutations of n items: \[ n! = n \times (n-1) \times (n-2) \times ... \times 1 \] For our 6 books, n is 6.
4Step 4: Calculate 6!
Now we calculate the factorial of 6: \[ 6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 \] Calculating this gives us: \[ 6 \times 5 = 30 \] \[ 30 \times 4 = 120 \] \[ 120 \times 3 = 360 \] \[ 360 \times 2 = 720 \] \[ 720 \times 1 = 720 \]
5Step 5: Conclusion
There are 720 different ways to arrange the 6 books on a shelf.
Key Concepts
FactorialCombinatoricsArrangements of Objects
Factorial
The concept of factorial is fundamental in many areas of mathematics and science. When we talk about the factorial of a number, we are referring to the product of that number and all the positive whole numbers less than it.
For example, to find the factorial of 6 (denoted as 6!), you would calculate:
Factorials grow very quickly, which is why they are useful in calculating permutations and combinations, as they represent large sets of possibilities and arrangements.
For example, to find the factorial of 6 (denoted as 6!), you would calculate:
- 6 × 5 × 4 × 3 × 2 × 1 = 720
Factorials grow very quickly, which is why they are useful in calculating permutations and combinations, as they represent large sets of possibilities and arrangements.
Combinatorics
Combinatorics is a branch of mathematics dealing with the counting, arrangement, and combination of objects. It provides a systematic way to analyze and count ways in which events can occur.
In our exercise with the books, combinatorics helps determine how many ways we can arrange them on a shelf. This is a permutation problem because the order of the books matters.
Combinatorics covers a range of concepts, including:
In our exercise with the books, combinatorics helps determine how many ways we can arrange them on a shelf. This is a permutation problem because the order of the books matters.
Combinatorics covers a range of concepts, including:
- Permutations: Ways to arrange objects where order is important.
- Combinations: Ways to select objects where order does not matter.
- Partitions: Ways to divide objects into distinct groups.
Arrangements of Objects
Arrangements of objects is a key aspect of permutation problems like the one with our 6 books on a shelf. When arranging objects, it is crucial to consider if the order matters.
If the order is important, as in our case, we use permutations to calculate the total number of different arrangements. The formula for permutations when arranging 'n' distinct objects is:
In scenarios where objects are considered indistinguishable in certain arrangements, modifications to the calculations are required, which often leads to the exploration of permutations with repetition or combinations.
Understanding the type of arrangement problem can inform the correct mathematical approach to find the solution."
If the order is important, as in our case, we use permutations to calculate the total number of different arrangements. The formula for permutations when arranging 'n' distinct objects is:
- \[ n! = n \times (n-1) \times (n-2) \times \ldots \times 1 \]
In scenarios where objects are considered indistinguishable in certain arrangements, modifications to the calculations are required, which often leads to the exploration of permutations with repetition or combinations.
Understanding the type of arrangement problem can inform the correct mathematical approach to find the solution."
Other exercises in this chapter
Problem 42
Find the probability of rolling a die five times and obtaining a 6 on the first two rolls, a 5 on the third roll, and a \(1,2,3,\) or 4 on the last two rolls.
View solution Problem 42
Find a general term \(a_{n}\) for the arithmetic sequence. $$a_{1}=-3, d=5$$
View solution Problem 43
Use Pascal's triangle to help expand the expression. $$ \left(2 x^{3}-y^{2}\right)^{3} $$
View solution Problem 43
Find a general term \(a_{n}\) for the arithmetic sequence. $$a_{3}=1, d=3$$
View solution