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Learn Calculation of Permutations and Combinations with InfinityLearn.com
Permutations and combinations are mathematical concepts that are used to calculate the number of ways that a set of objects can be arranged or selected. These concepts are used in a variety of fields, including mathematics, computer science, and statistics.
Permutations:
A permutation is an arrangement of a set of objects in a specific order. The number of permutations of a set of n objects is calculated using the formula:
n!/(n1! * n2! * … * nk!)
where n! is the factorial of n (n! = n * (n-1) * (n-2) * … * 2 * 1), and n1, n2, …, nk are the number of times that each object appears in the set.
For example, consider a set of three objects: {A, B, C}. The number of permutations of this set is 3!/(1! * 1! * 1!) = 3!/1! = 3!/1 = 3. The three permutations are:
{A, B, C} {A, C, B} {B, A, C} {B, C, A} {C, A, B} {C, B, A}
Combinations:
A combination is a selection of a set of objects, without regard to the order in which they are selected. The number of combinations of a set of n objects taken k at a time is calculated using the formula:
n!/(k! * (n-k)!)
For example, consider a set of three objects: {A, B, C}. The number of combinations of this set taken 2 at a time is 3!/(2! * (3-2)!) = 3!/2! = 3. The three combinations are:
{A, B} {A, C} {B, C}
Definition of Permutation and Combination
A permutation is an arrangement of a set of objects in a specific order. For example, the set {A, B, C} has six permutations: {A, B, C}, {A, C, B}, {B, A, C}, {B, C, A}, {C, A, B}, and {C, B, A}.
A combination is a selection of a set of objects, without regard to the order in which they are selected. For example, the set {A, B, C} has three combinations when taken two at a time: {A, B}, {A, C}, and {B, C}.
Permutations and combinations are used to calculate the number of ways that a set of objects can be arranged or selected. These concepts are useful in a variety of fields, including mathematics, computer science, and statistics.
The Difference between Permutations and Combinations
The main difference between permutation and combination is the order in which the objects are arranged or selected.
Permutation:
In a permutation, the order of the objects is important. For example, the set {A, B, C} has six permutations: {A, B, C}, {A, C, B}, {B, A, C}, {B, C, A}, {C, A, B}, and {C, B, A}.
Combination:
In a combination, the order of the objects is not important. For example, the set {A, B, C} has three combinations when taken two at a time: {A, B}, {A, C}, and {B, C}. These three combinations are considered to be different from each other, even though they contain the same objects.
Permutations and combinations are used to calculate the number of ways that a set of objects can be arranged or selected. These concepts are useful in a variety of fields, including mathematics, computer science, and statistics.
Basic Formula :
Permutation:
The basic formula for calculating the number of permutations of a set of n objects is:
n!/(n1! * n2! * … * nk!)
where n! is the factorial of n (n! = n * (n-1) * (n-2) * … * 2 * 1), and n1, n2, …, nk are the number of times that each object appears in the set.
For example, consider a set of three objects: {A, B, C}. The number of permutations of this set is 3!/(1! * 1! * 1!) = 3!/1! = 3!/1 = 3. The three permutations are:
{A, B, C} {A, C, B} {B, A, C} {B, C, A} {C, A, B} {C, B, A}
Combination:
The basic formula for calculating the number of combinations of a set of n objects taken k at a time is:
n!/(k! * (n-k)!)
For example, consider a set of three objects: {A, B, C}. The number of combinations of this set taken 2 at a time is 3!/(2! * (3-2)!) = 3!/2! = 3. The three combinations are:
{A, B} {A, C} {B, C}
Permutation and Combination Word Problems
Here are five permutation and combination word problems:
- You have a set of three books: {A, B, C}. In how many ways can you arrange the books on a shelf?
Solution: This is a permutation problem because the order in which the books are arranged is important. The number of permutations of a set of three objects is 3!/(1! * 1! * 1!) = 3!/1! = 3!/1 = 3. The three permutations are: {A, B, C}, {A, C, B}, and {B, A, C}.
- You have a set of four marbles: {R, G, B, W}. In how many ways can you draw three marbles from the set, without replacement?
Solution: This is a combination problem because the order in which the marbles are drawn is not important. The number of combinations of a set of four objects taken three at a time is 4!/(3! * (4-3)!) = 4!/3! = 4. The four combinations are: {R, G, B}, {R, G, W}, {R, B, W}, and {G, B, W}.
- You have a set of five cards: {A, 2, 3, 4, 5}. In how many ways can you draw two cards from the set, without replacement?
Solution: This is a combination problem because the order in which the cards are drawn is not important. The number of combinations of a set of five objects taken two at a time is 5!/(2! * (5-2)!) = 5!/2! = 10. The ten combinations are: {A, 2}, {A, 3}, {A, 4}, {A, 5}, {2, 3}, {2, 4}, {2, 5}, {3, 4}, {3, 5}, and {4, 5}.
- You have a set of six dice: {1, 2, 3, 4, 5, 6}. In how many ways can you roll the dice and get a sum of 9?
Solution: This is a permutation problem because the order in which the dice are rolled is important. The possible ways to roll the dice to get a sum of 9 are: (6, 3), (5, 4), and (4, 5). There are 3 permutations in total.
