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Permutations and Combinations

VIKAS VERMA
permutation and combination

Permutations and Combinations in which Permutations are the different arrangements or ways in which a set of items can be ordered, while combinations are the different ways in which a set of items can be selected without regard for order.

What is Permutation?

Permutation is a mathematical concept that deals with counting the number of different ways that a set of items can be arranged in a specific order. It is used to determine the total number of possible arrangements of a given set of items. It is often represented using the notation “nPk” where “n” is the total number of items and “k” is the number of items to be arranged. The formula for permutations is:

nPk = n! / (n-k)!

where “n!” is the factorial of n (n factorial), and “k!” is the factorial of k.

For example, if you have 3 items (A, B, and C) and you want to find the number of ways they can be arranged in a specific order, the number of permutations is 3! (3 factorial) which is 6. The permutations are: ABC, ACB, BAC, BCA, CAB, CBA.

It is important to note that permutations are different from combinations, which deal with counting the number of different ways that a set of items can be combined without regard to order.

What is a Combination?

Combination is a mathematical concept that deals with counting the number of different ways that a set of items can be combined without regard to order. It is used to determine the total number of possible subsets of a given set of items. It is often represented using the notation “nCk” where “n” is the total number of items and “k” is the number of items to be combined. The formula for combinations is:

nCk = n! / (k! * (n-k)!)

where “n!” is the factorial of n (n factorial), “k!” is the factorial of k, and “(n-k)!” is the factorial of (n-k).

For example, if you have 3 items (A, B, and C) and you want to find the number of ways they can be combined without regard to order, the number of combinations is 3C2, which is 3. The combinations are: AB, AC, BC.

It is important to note that combinations are different from permutations, which deal with counting the number of different ways that a set of items can be arranged in a specific order.

In many cases, the real-world problems can be solved more efficiently by using combinations rather than permutations.

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Permutation
Combination


Difference Between Permutation and Combination

Permutation and combination are both mathematical concepts that deal with counting the number of different ways that a set of items can be arranged or combined. However, they differ in how they treat the order of the items:

Permutation:

Combination:

For example, if you have 3 items (A, B, and C) and you want to find the number of ways they can be arranged in a specific order, the number of permutations is 3! (3 factorial) which is 6. The permutations are: ABC, ACB, BAC, BCA, CAB, CBA. However, if you want to find the number of ways they can be combined without regard to order, the number of combinations is 3C2, which is 3. The combinations are: AB, AC, BC.

PermutationCombination
Arranging people, digits, numbers, alphabets, letters, and coloursSelection of menu, food, clothes, subjects, team.
Picking a team captain, pitcher and shortstop from a group.Picking three team members from a group.
Picking two favourite colours, in order, from a colour brochure.Picking two colours from a colour brochure.
Picking first, second and third place winners.Picking three winners.
Permutation FormulaCombination Formula
nPr = (n!) / (n-r)!

Permutations:

Example 1: A committee of 4 members is to be formed from a group of 8 people. How many ways can this be done? Solution: The number of ways to form a committee of 4 members from a group of 8 people is 8P4 = 8! / (8-4)! = 8! / 4! = (8 x 7 x 6 x 5) / (4 x 3 x 2 x 1) = 70.

Example 2: In how many ways can the letters of the word “LEADER” be arranged? Solution: The number of ways the letters of the word “LEADER” can be arranged is 6!.

Combinations:

Example 1: A committee of 4 members is to be selected from a group of 8 people. How many ways can this be done? Solution: The number of ways to select a committee of 4 members from a group of 8 people is 8C4 = 8! / (4! x 4!) = 70 / (4 x 3 x 2 x 1) = 35.

Example 2: In how many ways can 4 letters be selected from the word “LEADER”? Solution: The number of ways to select 4 letters from the word “LEADER” is 6C4 = 6! / (4! x 2!) = 15.

Note that the difference between permutations and combinations is that permutations take into account the order of elements, while combinations do not.

Uses of Permutation and Combination

Permutations and combinations have many applications in a wide range of fields, including mathematics, statistics, computer science, physics, and engineering. Here are a few examples of how permutations and combinations are used:

  1. Combinatorics: Permutations and combinations are used in combinatorics, the branch of mathematics that deals with counting and arranging objects. They are used to determine the number of ways that a set of items can be arranged or combined.
  2. Probability: Permutations and combinations are used in probability theory to calculate the probability of different outcomes in random experiments. For example, in a game of poker, permutations and combinations can be used to determine the probability of getting a specific hand.
  3. Cryptography: Permutations and combinations are used in cryptography to create and analyze encryption algorithms.
  4. Computer Science: Permutations and combinations are used in computer science to solve problems like the traveling salesman problem and other optimization problems.
  5. Scheduling: Permutations and combinations are used in scheduling to determine the possible ways to schedule a set of tasks.
  6. Physics: Permutations and combinations are used in physics to calculate the number of ways that a set of particles can be arranged in a specific configuration.
  7. Genetics: Permutations and combinations are used in genetics to calculate the number of possible genetic combinations.
  8. Marketing: Permutations and combinations are used in marketing to determine the number of possible ways to advertise a product.
  9. Business: Permutations and combinations are used in business to calculate the number of possible ways to arrange a set of products in a store.
  10. Daily life: Permutations and combinations are used in daily life when choosing a combination lock, making a schedule, and many other similar activities.

Permutation and Combination Formulas

There are several formulas that are commonly used to calculate permutations and combinations. Here are a few examples:

  1. Permutation of n distinct objects taken r at a time: nP r = n! / (n-r)!
  2. Combination of n distinct objects taken r at a time: nC r = n! / (r! * (n-r)!)
  3. Permutation of n objects where m are similar and the rest are different: nP r = n! / (n-r)! * m^r
  4. Combination of n objects where m are similar and the rest are different: nC r = nC r = (n+m-1)C r = (n+m-1)C (n-r) = (n+m-1)! / (r! * (n-r)! * (m-1)!)
  5. Permutation of n objects with repetition: P(n,r) = n^r
  6. Combination of n objects with repetition: C(n,r) = (n+r-1)C r = (n+r-1)! / (r! * (n-1)!)
  7. Permutation of n objects in a circle: nP n = n!
  8. Permutation of n objects with restrictions: nP r = n! / (n-r)! * r! (n1-1)! (n2-1)! …. (nr-1)! where n1, n2, …. nr are the restrictions.

Note that in the above formulas n! denotes the factorial of n, which is the product of all positive integers from 1 to n.

Problems of Permutations and Combinations for students

Permutations and combinations are mathematical concepts that can be used to solve a wide range of problems in various fields such as combinatorics, probability, statistics, computer science, and physics. However, there are some problems that can arise when working with these concepts:

  1. Difficulty in understanding the difference between permutations and combinations: As both concepts deal with counting the number of different ways that a set of items can be arranged or combined, it can be difficult for some people to understand the difference between the two and when to use one versus the other.
  2. Complexity of calculations: Permutations and combinations can involve large numbers and factorials, which can make calculations difficult and time-consuming.
  3. Difficulty in applying the concepts to real-world problems: Even though permutations and combinations are useful concepts, they can be difficult to apply to real-world problems, especially when the problem is not well defined or when the data is incomplete or unreliable.
  4. Misinterpretation of results: It is important to interpret the results correctly, as the results of permutations and combinations are often given in terms of the number of possible outcomes, rather than the probability of a specific outcome, so it is important to carefully interpret the results.
  5. Difficulty in understanding the use of permutations and combinations in different fields: The use of permutations and combinations may vary between different fields, so it can be difficult for some people to understand how these concepts can be applied in different fields such as physics, cryptography, and business.

It is important to keep in mind that these problems can be overcome with practice and a good understanding of the concepts.

Solved Examples of permutations and combinations for class 9

Here are a few examples of solving problems involving permutations and combinations for class 9:

  1. In a class of 25 students, a monitor is to be selected. How many ways can this be done? Solution: In this case, we are choosing 1 student (r) out of 25 students (n) without replacement and without order. So the formula to use is nP r = n! / (n-r)! = 25P1 = 25. So there are 25 ways to select the monitor.
  2. A cricket team is to be selected from a group of 15 players. How many ways can this be done if the team has to have exactly 11 players? Solution: In this case, we are choosing 11 players (r) out of 15 players (n) without replacement and without order. So the formula to use is nC r = n! / (r! * (n-r)!) = 15C11 = 1365. So there are 1365 ways to select the team of 11 players.
  3. A lottery is to be held with 10 numbers between 1 and 20. In how many ways can the numbers be chosen? Solution: In this case, we are choosing 10 numbers (r) out of 20 numbers (n) without replacement and without order. So the formula to use is nC r = n! / (r! * (n-r)!) = 20C10 = 184756. So there are 184756 ways to choose the numbers in the lottery.
  4. A group of 7 friends is to go on a trip together. In how many ways can they sit in a car with 7 seats? Solution: In this case, we are choosing 7 friends (r) out of 7 friends (n) without replacement and with order. So the formula to use is nP r = n! / (n-r)! = 7P7 = 5040. So there are 5040 ways for them to sit in the car.
  5. A code is to be made with 5 characters where each character is either a letter or a digit. How many possible codes can be created? Solution: In this case, we are choosing 5 characters (r) out of 26 letters and 10 digits (n) with replacement and with order. So the formula to use is P(n,r) = n^r = (26+10)^5 = 36^5 = 2985984. So there are 2985984 possible codes.

Solved Examples of permutations and combinations for class 10

Here are a few examples of solving problems involving permutations and combinations for class 10:

  1. In how many ways can a committee of 5 be formed from a group of 8 men and 6 women? Solution: In this case, we are choosing 5 people (r) out of 8 men and 6 women (n) without replacement and without order. So the formula to use is nC r = n! / (r! * (n-r)!). The number of ways to choose 5 people from 8 men = 8C5 = 56. The number of ways to choose 5 people from 6 women = 6C5 = 6. The total number of ways = 56 * 6 = 336.
  2. A box contains 8 red, 7 blue and 5 yellow balls. In how many ways can 3 balls be drawn from the box? Solution: In this case, we are choosing 3 balls (r) out of 8 red, 7 blue and 5 yellow balls (n) without replacement and without order. So the formula to use is nC r = n! / (r! * (n-r)!). The total number of balls is 8+7+5 = 20. So the total number of ways to choose 3 balls is 20C3 = 1140.
  3. A password must consist of 4 digits and 4 letters. In how many ways can a password be created? Solution: In this case, we are choosing 4 letters (r1) out of 26 letters (n1) and 4 digits (r2) out of 10 digits (n2) with replacement and with order. So the formula to use is P(n1,r1) * P(n2,r2) = 26^4 * 10^4 = 4569760000. So there are 456,976,000 possible passwords.
  4. A club has 8 members. In how many ways can a president, vice-president and secretary be chosen from the members? Solution: In this case, we are choosing 3 people (president, vice-president and secretary) out of 8 members (n) without replacement and without order. So the formula to use is nP r = n! / (n-r)! = 8P3 = 56. So there are 56 ways to choose the president, vice-president, and secretary.
  5. A bag contains 4 white, 3 red and 2 blue balls. In how many ways can 3 balls be drawn from the bag such that at least one ball is blue? Solution: In this case, we are choosing 3 balls (r) out of 4 white, 3 red and 2 blue balls (n) without replacement and without order. So the formula to use is nC r = n! / (r! * (n-r)!). The total number of ways to choose 3 balls is (4+3+2)C3 = 9C3 = 84. But we need to subtract the ways of choosing 3 balls without blue ball, which is (4+3)C3 = 7C3 = 35. So, the required number of ways is 84-35 = 49

Solved Examples of permutations and combinations for class 11

Here are a few examples of solving problems involving permutations and combinations for class 11:

  1. A test consists of 10 multiple choice questions, each having 4 options. In how many ways can a student attempt all questions?
    • Solution: In this case, we are choosing an option (r) out of 4 options (n) for each of the 10 questions (n’) with replacement and without order. So the formula to use is P(n,r)^n’ = 4^10 = 1048576. So there are 1,048,576 ways for a student to attempt all questions.
  2. A box contains 5 white, 7 black and 6 red balls. In how many ways can 4 balls be drawn from the box such that at least one ball is white?
    • Solution: In this case, we are choosing 4 balls (r) out of 5 white, 7 black and 6 red balls (n) without replacement and without order. So the formula to use is nC r = n! / (r! * (n-r)!). The total number of ways to choose 4 balls is (5+7+6)C4 = 18C4 = 18564. But we need to subtract the ways of choosing 4 balls without white ball, which is (7+6)C4 = 13C4 = 715. So, the required number of ways is 18564-715 = 17849
  3. A committee of 6 members is to be formed from a group of 8 men and 5 women. In how many ways can the committee be formed if it must have at least one woman?
    • Solution: In this case, we are choosing 6 members (r) out of 8 men and 5 women (n) without replacement and without order. So the formula to use is nC r = n! / (r! * (n-r)!). The total number of ways to choose 6 members is (8+5)C6 = 13C6 = 715. But we need to subtract the ways of choosing 6 members without a woman, which is 8C6 = 56. So, the required number of ways is 715-56 = 659
  4. A password must consist of 2 letters and 3 digits. In how many ways can a password be created if repetition of letters or digits is not allowed?
    • Solution: In this case, we are choosing 2 letters (r1) out of 26 letters (n1) and 3 digits (r2) out of 10 digits (n2) without replacement and with order. So the formula to use is n1P r1 * n2P r2 = 26P2 * 10P3 = 26251098 = 353800. So there are 353,800 possible passwords.
  5. A container has 6 red and 4 green balls. A sample of 3 balls is drawn from the container at random. Find the probability that the sample contains at least one green ball.
    • Solution: In this case, we are choosing 3 balls (r) out of 6 red and 4 green balls (n) without replacement and without order. So the formula to use is nC r = n! / (r! * (n-r)!). The total number of ways to choose 3 balls is (6+4)C3 = 10C3 = 120. The number of ways to choose 3 green balls is 4C3 = 4. The number of ways to choose 2 green and 1 red balls is 4C2 * 6C1 = 24. So the probability that the sample contains at least one green ball is (4+24)/120 = 0.4 or 40%.

Solved Examples of permutations and combinations for class 12

Permutations:

Example 1: A professor has to grade 12 papers from 3 students. In how many ways can she do this? Solution: The professor has 3 students to grade papers from, and each student has 12 papers, so there are a total of 36 options for each paper. The total number of ways the professor can grade the papers is 36P12 = 36! / (36-12)! = 9,958,869,056,000.

Example 2: A club has 10 members, and the president has to choose a committee of 4 members to plan an event. In how many ways can the president choose the committee? Solution: The president has 10 members to choose from, so there are a total of 10 options for each of the 4 members on the committee. The total number of ways the president can choose the committee is 10P4 = 10! / (10-4)! = 210.

Combinations:

Example 1: A professor has to grade 12 papers from 3 students. In how many ways can she do this without caring about the order? Solution: The professor has 3 students to grade papers from, and each student has 12 papers, so there are a total of 36 options for each paper. The total number of ways the professor can grade the papers without caring about the order is 36C12 = 36! / (12! x 24!) = 7,890.

Example 2: A club has 10 members, and the president has to choose a committee of 4 members to plan an event. In how many ways can the president choose the committee without caring about the order? Solution: The president has 10 members to choose from, so there are a total of 10 options for each of the 4 members on the committee. The total number of ways the president can choose the committee without caring about the order is 10C4 = 210.

Example 3: In how many ways can a group of 10 people be divided into two teams of 5? Solution: The number of ways to divide 10 people into two teams of 5 is 10C5 = 252.

Example 4:A group of 10 people are to be seated in a row. In how many ways can this be done? Solution: The number of ways to seat 10 people in a row is 10!

Example 5: A group of 8 people are to be seated in a circle. In how many ways can this be done? Solution: The number of ways to seat 8 people in a circle is (8-1)! = 7!

Permutations and Combinations in hindi

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Permutation and combination concepts

Permutation is the arrangement of objects in a specific order, while combination is the selection of objects without regard to the order in which they are arranged. In other words, permutation takes into account the order of the objects, while combination does not.

For example, if you have a set of three letters {A,B,C}, a permutation would be the different ways you can arrange those letters in a specific order, such as ABC, BAC, CAB. A combination would be the different ways you can select two letters without regard to the order, such as AB or BC.

The formulas for calculating the number of permutations and combinations are different. The formula for permutations of n objects taken r at a time is n!/(n-r)! and the formula for combination of n objects taken r at a time is n!/(r!(n-r)!).

permutation and combination difficult questions with answers

  1. How many different ways can the letters of the word “MATHEMATICS” be arranged? Answer: The word “MATHEMATICS” has 10 letters, so the number of ways to arrange them is 10! (10 factorial) which is 3,628,800.
  2. A committee of 5 people is to be chosen from a group of 8 people. How many different committees are possible? Answer: The number of ways to choose 5 people from a group of 8 is 8!/(5!3!) = 56.
  3. A locker combination is made up of 3 digits. How many different combinations are possible if the digits can range from 0 to 9? Answer: There are 10 possible choices for each digit (0-9), so there are 101010 = 1000 possible combinations.
  4. A password consists of 4 letters and 2 digits. How many different passwords can be formed if repetition of letters and digits is not allowed? Answer: The number of ways to choose 4 letters from a group of 26 letters is 26!/(4!22!) = 26252423 = 358,800. The number of ways to choose 2 digits from a group of 10 digits is 10!/(2!8!) = 45. So the total number of different passwords is 358,80045 = 16,130,000.
  5. A card deck contains 10 red cards and 8 black cards. A player draws 4 cards at random. How many different ways can 4 cards be drawn such that there are 2 red and 2 black cards in the drawn cards? Answer: The number of ways to draw 2 red cards from 10 red cards is 10C2 = 45. The number of ways to draw 2 black cards from 8 black cards is 8C2 = 28. The total number of ways to draw 2 red and 2 black cards is 45*28 = 1260

Note: C stands for combination and denotes nCr = n! / (r! * (n-r)!)

These are some examples of permutation and combination problems, but keep in mind that there are many variations and different levels of difficulty. It’s important to practice solving a wide range of problems to develop a strong understanding of these concepts.

Permutation and combination class 12 exercise

  1. A class of 12 students is to be divided into two teams of 6 each. How many different ways can this be done? Answer: The number of ways to divide 12 students into two teams of 6 is 12C6 = 924.
  2. A group of 8 people is to be seated at a round table. In how many ways can this be done? Answer: The number of ways to seat 8 people around a round table is 8P8 = 8!, which is 40320.
  3. A box contains 4 red, 5 blue, and 6 green balls. A sample of 3 balls is to be drawn without replacement. In how many ways can this be done? Answer: The number of ways to choose 3 balls from a group of 4 red, 5 blue, and 6 green balls is 15C3 = 455.
  4. A group of 5 people is to be selected from a group of 8 men and 12 women. In how many ways can this be done if the group must contain at least one woman? Answer: The number of ways to select 5 people from a group of 8 men and 12 women is 20C5 = 15504. Subtract the number of ways to select a group of 5 people without a woman, which is 8C5 = 56, from the total number of ways. 15504 – 56 = 15448.
  5. A bag contains 4 white, 5 black and 6 brown balls. A sample of 3 balls is to be drawn with replacement. In how many ways can this be done? Answer: There are 15 balls in total, so there are 151515 = 3,375 possible ways to draw 3 balls with replacement.
  6. A committee of 4 members is to be selected from a group of 7 men and 5 women. In how many ways can this be done if the committee must contain at least one woman? Answer: The number of ways to select 4 people from a group of 7 men and 5 women is 12C4 = 495. Subtract the number of ways to select a committee of 4 people without a woman, which is 7C4 = 35, from the total number of ways. 495 – 35 = 460

Keep in mind that the above are just examples of permutation and combination problems, and there are many variations and different levels of difficulty. It’s important to practice solving a wide range of problems to develop a strong understanding of these concepts.

Frequently Asked Questions on Permutations and Combinations

What do you mean by permutations and combinations?

A permutation is an act of arranging objects or numbers in order.
Combinations are the way of selecting objects or numbers from a group of objects or collections, in such a way that the order of the objects does not matter.

Give examples of permutations and combinations.

An example of permutations is the number of 2 letter words that can be formed by using the letters in a word say, GREAT; 5P_2 = 5!/(5-2)!
An example of combinations is in how many combinations we can write the words using the vowels of the word GREAT; 5C_2 =5!/[2! (5-2)!]

What is the formula for permutations and combinations?

The formula for permutations is: nPr = n!/(n-r)!
The formula for combinations is: nCr = n!/[r! (n-r)!]

What are the real-life examples of permutations and combinations?

Arranging people, digits, numbers, alphabets, letters, and colours are examples of permutations.
Selection of menu, food, clothes, subjects, the team are examples of combinations.

Write the relation between permutations and combinations.

The formula for permutations and combinations are related as:
nCr = nPr/r!

Give the applications of permutation and combination in mathematics.

In Mathematics, the concept called “permutation and combinations” are applied in probability, relations and functions, set theory and so on.

What is the factorial formula?

The factorial formula is used in the calculation of permutations and combinations, which is obtained by taking the product of all numbers in the sequence (i.e., from 1 to n). For example, 3! = 3 × 2 × 1 = 6.

What does nCr represent?

nCr represents the number of combinations from “n” objects taken “r” at a time.

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