Permutations and combinations | Complete overview with problem and solution

Permutations and Combinations

Illustration of the concept Permutations and Combinations by the examples

In this article, we have discussed some examples which will make the foundation strong of the students on Permutations and Combinations to get the insight clearance of the concept, it is well aware  that the Permutations and combinations both are the process to calculate the possibilities, the difference between them is whether order matters or not, so here by going through the number of examples we will get clear the confusion where to use which one.

The methods of arranging or selecting a small or equal number of people or items at a time from a group of people or items provided with due consideration to be arranged in order of planning or selection are called permutations.

Each different group or selection that can be created by taking some or all of the items, no matter how they are organized, is called a combination.

Basic Permutation (nPr formula) Examples

            Here We are making group of n different objects, selected r at a time equivalent to filling r places from n things.

Combinations
Permutations and Combinations

The number of ways of arranging = The number of ways of filling r places.

^{n}P_{r}=n.(n-1).(n-2)…(n-r+1)=\frac{n!}{(n-r)!}

\frac{n.(n-1).(n-2)…(n-r+1).(n-r)!}{(n-r)!}=\frac{n!}{(n-r)!}=^{n}P_{r}

so nPr formula we have to use is

^{n}P_{r}=\frac{n!}{(n-r)!}

Example 1): There is a train whose 7 seats are kept empty, then how many ways can three passengers sit.

solution: Here n=7, r=3

so      Required number of ways=

^{n}P_{r}=\frac{n!}{(n-r)!}

^{7}P_{3}=\frac{7!}{(7-3)!}=\frac{4!.5.6.7}{4!} =210

In 210 ways 3 passengers can sit.

Example 2) How many ways can 4 people out of 10 women be chosen as team leaders?

solution: Here n=10, r=4

so      Required number of ways=

^{n}P_{r}=\frac{n!}{(n-r)!}

^{10}P_{4}=\frac{10!}{(10-4)!}=\frac{6!.7.8.9.10}{6!}=5040

In 5040 ways 4 women can be chosen as team leaders.

Example 3) How many permutations are possible from 4 different letter, selected from the twenty-six letters of the alphabet?

solution: Here n=26, r=4

so      Required number of ways=

^{n}P_{r}=\frac{n!}{(n-r)!}

^{26}P_{4}=\frac{26!}{(26-4)!}=\frac{22!.23.24.25.26}{22!}=358800

In 358800 ways, 4 different letter permutations are available.

Example 4) How many different three-digit permutations are available, selected from ten digits from 0 to 9 combined?(including 0 and 9).

solution: Here n=10, r=3

so      Required number of ways=

^{n}P_{r}=\frac{n!}{(n-r)!}

^{10}P_{3}=\frac{10!}{(10-3)!}=\frac{7!.8.9.10}{7!}=720

In 720 ways, three-digit permutations are available.

Example 5) Find out the number of ways a judge can award a first, second, and third place in a contest with 18 competitors.

solution: Here n=18, r=3

so      Required number of ways=

^{n}P_{r}=\frac{n!}{(n-r)!}

^{18}P_{3}=\frac{18!}{(18-3)!}=\frac{15!.16.17.18}{15!}=4896

Among the 18 contestants, in 4896 number of ways, a judge can award a 1st, 2nd and 3rd place in a contest.

Example

6) Find the number of ways, 7 people can organize themselves in a row.

solution: Here n=7, r=7

so      Required number of ways=

^{n}P_{r}=\frac{n!}{(n-r)!}

^{7}P_{7}=\frac{7!}{(7-7)!}=\frac{7!}{0!}=5040

In 5040 number of ways, 7 people can organize themselves in a row.

Examples based on Combination (nCr formula/ n choose k formula)

The number of combinations (selections or groups) that can be set up from n different objects taken r (0<=r<=n) at a time is

^{n}C_{r}=\frac{n!}{r!(n-r)!}=\frac{^{n}P_{r}}{r!} =\binom{n}{k}

This is commonly known as nCr or n choose k formula.

^{n}C_{k}=\frac{n!}{k!(n-k)!}

Examples:

1) If you have three dress with different colour in red, yellow and white then can you find a different combination you get if you have to choose any two of them?

Solution: here n=3, r=2 this is 3 CHOOSE 2 problem

^{n}C_{r}=\frac{n!}{r!(n-r)!}

^{3}C_{2}=\frac{3!}{2!(3-2)!}=\frac{2!.3}{2!.1}=3

In 3 different combination you get any two of them.

2) How many different combinations can be done if you have 4 different items and you have to choose 2?

Solution: here n=4, r=2 this is 4 CHOOSE 2 problem

^{n}C_{r}=\frac{n!}{r!(n-r)!}

^{4}C_{2}=\frac{4!}{2!(4-2)!}=\frac{2!.3.4}{2!.2!}=6

In 6 different combination you get any two of them.

3) How many different combinations can be made if you have only 5 characters and you have to choose any 2 among them?

Solution: here n=5, r=2 this is 5 CHOOSE 2 problem

^{n}C_{r}=\frac{n!}{r!(n-r)!}

^{5}C_{2}=\frac{5!}{2!(5-2)!}=\frac{3!.4.5}{2!.3!}=10

In 10 different combination you get any two of them.

4) Find the number of combinations 6 choose 2.

Solution: here n=6, r=2 this is 6 CHOOSE 2 problem

^{n}C_{r}=\frac{n!}{r!(n-r)!}

^{6}C_{2}=\frac{6!}{2!(6-2)!}=\frac{4!.5.6}{2!.4!}=15

In 15 different combination you get any two of them.

5) Find the number of ways of choosing 3 members from 5 different partners.

Solution: here n=5, r=3 this is 5 CHOOSE 3 problem

^{n}C_{r}=\frac{n!}{r!(n-r)!}

^{5}C_{3}=\frac{5!}{3!(5-3)!}=\frac{3!.4.5}{3!.2!}=10

In 10 different combination you get any three of them.

6) Box of crayons having red, blue, yellow, orange, green and purple. How many unlike ways can you use to draw only three colour?

Solution: here n=6, r=3 this is 6 CHOOSE 3 problem

^{n}C_{r}=\frac{n!}{r!(n-r)!}

^{6}C_{3}=\frac{6!}{3!(6-3)!}=\frac{3!.4.5.6}{3!.3.2.1}=20

In 20 different combination you get any three of them.

7) Find the number of combinations for 4 choose 3.

Solution: here n=4, r=3 this is 4 CHOOSE 3 problem

^{n}C_{r}=\frac{n!}{r!(n-r)!}

^{4}C_{3}=\frac{4!}{3!(4-3)!}=\frac{3!.4}{3!.1!}=4

In 4 different combination you get any three of them.

8) How many different five-person committees can be elected from 10 people?

Solution: here n=10, r=5 this is 10 CHOOSE 5 problems

^{n}C_{r}=\frac{n!}{r!(n-r)!}

^{10}C_{5}=\frac{10!}{5!(10-5)!}=\frac{10!}{5!.5!}=\frac{5!.6.7.8.9.10}{5!.5.4.3.2}=7.4.9=252

So 252 different 5 person committees can be elected from 10 people.

9) There are 12 volleyball players in total in college, which will be made up of a team of 9 players. If the captain remains consistent, the team can be formed in how many ways.

Solution: here as the captain already has been selected, so now among 11 players 8 are to be chosen n=11, r=8 this is 11 CHOOSE 8 problem

^{n}C_{r}=\frac{n!}{r!(n-r)!}

^{11}C_{8}=\frac{11!}{8!(11-8)!}=\frac{11!}{8!.3!}=\frac{8!.9.10.11}{8!.3.2.1}=3.5.11=165

So If the captain remains consistent, the team can be formed in 165 ways.

10) Find the number of combinations 10 choose 2.

Solution: here n=10, r=2 this is 10 CHOOSE 2 problem

^{n}C_{r}=\frac{n!}{r!(n-r)!}

^{10}C_{2}=\frac{10!}{2!(10-2)!}=\frac{10!}{2!.8!}=\frac{8!.9.10}{2!.8!}=5.9=45

In 45 different combination you get any two of them.

We have to see the difference that nCr is the number of ways things can be selected in ways r and nPr is the number of ways things can be sorted by means of r. We have to keep in mind that for any case of permutation scenario, the way things are arranged is very very important. However, in Combination, the order means nothing.

Conclusion

A detailed description with examples of the Permutations and combinations has been provided in this article with few real-life examples, in a series of articles we will discuss in detail the various outcomes and formulas with relevant examples if you are interested in further study go through this link.

Reference

  1. SCHAUM’S OUTLINE OF Theory and Problems of DISCRETE MATHEMATICS
  2. https://en.wikipedia.org/wiki/Permutation
  3. https://en.wikipedia.org/wiki/Combination

About DR. MOHAMMED MAZHAR UL HAQUE

I am DR. Mohammed Mazhar Ul Haque , Assistant professor in Mathematics. Having 12 years of experience in teaching. Having vast knowledge in Pure Mathematics , precisely on Algebra. Having the immense ability of problem designing and solving. Capable of Motivating candidates to enhance their performance.
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