Permutations and Combinations, Study notes of Mathematics

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PERMUTATIONS

AND

COMBINATIONS

[MADE EASY]

BY

DANIEL O. AMAIZE

With

[Standard questions (including GMAT) and solutions]

TABLE OF CONTENT

• Cover page ----------------------------------------
• Table of content ----------------------------------------
• Introduction ----------------------------------------
• The keyword approach to permutations ----------------------------
• The keyword approach to combinations ----------------------------
• questions -------------------------------------- Visualizing the difference between permutation and combination
• Exercises --------------------------------------
• Solutions to exercises ---------------------------------------
• References ---------------------------------------
• Appendix --------------------------------------

and combination problems. Here we are going to look at the keyword approach to understanding permutation problems. Let us understand this keyword approach by solving two problems.

Problem 1

How many three-letter words can be formed from the alphabets A, B, C, and D?

Solution

The 3-letter words that can be formed from the letters A, B, C and D are:

ABC, ABD, ACB,

ACD, ADB, ADC,

BAC, BAD, BCD,

BDC, BCA, BDA,

CAB, CAD, CDA,

CBA, CBD, CDB,

DCA, DAC, DBA,

DAB, DCB, DBC

That’s a total of 24 different words or arrangements of the four letters. These arrangements are known as permutations.

Now let us use the keyword approach to solve this problem.

In this approach, we will first select the three letters and then arrange the selected letters using their orders:

Select A, B and C.

Select A, B and D

Select A, D and C

Select B, C and D

This can be done in 4 ways as shown above.

We now have three letters each from the 4 different anrrangements which we can arranged in 6 different ways:

ABC, ACB, BAC, BCA, CAB and CBA.

ABD, ADB, BAD, BDA, DAB and DBA.

ADC, ACD, CAD, CDA, DAC and DCA.

BCD, BDC, CBD, CDB, DCB and DBC.

Counting all the cases, we have a total of 24 ways or arrangements of the 4 letters taking 3 at a time. Thus, permutation involves selection and then arrangement of the different selections.

From our understanding, the formula to arrange n objects, taking r at a time is nPr = (^) (𝑛−𝑟)!𝑛!

Where n = number of items or objects to be arranged

r = number of items or objects taken at a time

n! = n(n – 1)(n – 2)…….. 2 * 1

Now let us apply our formula to solve the problem above:

n = 4 (A, B, C and D), r = 3 (3-letter words)

nPr = nPr = (^) (𝑛−𝑟)!𝑛! = 4P 3 =

4! (4−3)! =^

4 ∗ 3 ∗ 2 ∗1! 1! = 4 * 3 * 2 = 24

Before we take another example, let us look at some important keywords used in describing permutations. There are three main keywords used often to imply permutations in questions. Look out for:

 Arrangemment  Ordered  Unique

Example 2

In how many different ways can 12 books be arranged on a shelf if 3 books are to be on each strand?

Example 1

From four players A, B, C, and D, how many triples team can be formed?

Solution

Taking a first look at the question, it looks like the question of Example 1 on permutation right?

But believe me, they are different.

What we are required to do here is to select a group of 3 objects from four objects. So in how many ways can we do that?

Remember that we are dealing with a team of people and not letters. So ABC is the same as BAC, CAB, BCA, ACB and CBA here, because the order doesn’t matter when people are involved. This is a very good example of combination.

From 4 players A, B, C and D, the teams of 3-players can be:

Team ABC

Team ABD

Team ACD

Team BCD

Thus, we can have only four triples team from four players.

Now that we have solved the problem manually, let us take a look at the keyword approach.

We are going to list all cases of forming a triples team from 4 players as shown below:

Select A, B and C

Select A, B and D

Select A, C and D

Select B, C and D

That is a total of four ways. Now compare this with the solution to the problem of forming

3 - letter words from 4 letters.

Let the selection of r objects from n objects be denoted as nCr.

Then, nCr = 4 and nPr = 24

That is, nPr = 4 * 6 = nCr * 3! = nCr * r!

Thus, nCr =

𝑛𝑃𝑟 𝑟! =^ nPr =^

𝑛! (𝑛−𝑟)! ∗ 𝑟!

For our example above, n = 4 and r = 3

nCr =

𝑛! (𝑛−𝑟)! ∗ 𝑟! =^

4! ( 4 − 3 )! ∗ 3! =^

4! 1! ∗ 3! =^

4 ∗ 3! 1 ∗ 3! =^4

Before we take another example, let us take note of the following keywords which will aid us to identify problems in combination:

 Select  Choose  Pick  Combination

Example 2

A bowl contains 6 different fruits. In how many ways can we choose a pair of 2 different fruits from the bowl?

Solution

Since we are choosing items that are not identical, this is a selection and a combination problem.

Thus, n = 6 , r = 2

nCr = 6C 2 = 6! ( 6 − 2 )! ∗ 2! =^

6! 4! ∗ 2! =^

6 ∗ 5 ∗ 4! 4! ∗ 2! =^

6 ∗ 5 2 ∗ 1 =^15

5! (5−3)! ∗ 3! *^

6! (6−4)! ∗ 4! ∗^

3! (3−2)! ∗ 2!

=

5! 2! ∗ 3! *^

6! 2! ∗ 4! ∗^

3! 1! ∗ 2!

= 5 ∗ 4 ∗ 3! 2! ∗ 3! *^

6 ∗ 5 ∗ 4! 2! ∗ 4! ∗^

3 ∗ 2! 1! ∗ 2!

= 5 ∗ 4 2! *^

6 ∗ 5 2! ∗^

3 1!

= 10 * 15 * 3

Therefore, the number of different combinations of 9 stocks = 450 combinations

VISUALIZING THE DIFFERENCE BETWEEN PERMUTATION AND

COMBINATION QUESTIONS

Sometimes, none of the identifying keywords are used in a question. So how do we determine if such a question is on permutation or combination? Well, that is what this section is all about. In such cases, we try to visualize the difference between permutation and combination problems.

Consider the following examples:

Example 1

There are 18 teams in a certain league, and each team plays with the other teams exactly twice. What is the total number of games played in the league?

Solution

There are 18 teams in the league and we know that each game is played between two teams only.

 The match between team A and team B is the same as the match between team B and team A  Thus, for each match to happen, we must select 2 teams only and in this case, the arrangement will not matter

Can you observe we arrived at the keyword SELECT by carefully analyzing the information given, and making some meaningful inferences?

Now, we only have to find the number of ways of selecting 2 teams from 18 teams and then

multiply by 2 since there are two games (home and away) played between any two teams_._ Hence, total number of games played = 2! * 18C 2

= 2 * 18! ( 18 − 2 )! ∗ 2!

= 2 *

18! 16! ∗ 2!

= 2 * 18 ∗ 17 ∗ 16! 16! ∗ 2!

= 2 *

18 ∗ 17 2 = 18 * 17 Thus, the total number of games played = 306 games

Example 2 A company held a board meeting to appoint three out of their 10 board members for the post of CEO, DCEO and COO. In how many ways can 3 members of the board get the mandate of the company?

Solution Since there are three different positions and three persons required to fill those positions, each of the three persons selected can fill any of the three positions. Then, we can understand that the problem involves both selection and arrangement. Hence, it is a permutation problem. Thus, the number of ways 3 members can be selected from 10 members to fill 3 different

positions = 10P 3 = 10! ( 10 − 3 )! =^

10! 7! =^

10 ∗ 9 ∗ 8 ∗ 7! 7! = 10 * 9 * 8 Therefore, the total number of ways = 720 ways

After going through the explanations and example problems, you can try out the following exercises. Try to solve them first before going down to the solutions section to check if you have solved them correctly.

SOLUTION TO EXERCISES

1. This is a permutation problem since order is important with words Thus, n = 10, r = 4 nPr = 10P4 =

10! (10−4)! =^

10! 6! =^

10 ∗ 9 ∗ 8 ∗ 7 ∗ 6! 6! = 10 * 9 * 8 * 7 therefore, the number of 4-letter words = 5040 words

2. For n persons to seat on a round table, one person has to be used as a marker since a round table has no beginning or end. Then, the other (n – 1) persons can be seated in (n – 1)! Ways Thus the number of ways 8 men can be seated on a round table = (8 – 1)! Ways = 7! Ways = 7 * 6 * 5 * 4 * 3 * 2 * 1 Therefore, the number of ways 8 men can be seated on a round table = 5040 ways
3. The number of ways of selecting a pair of two different books from 9 books = 9C2 ways

=

9! (9−2)! 2! =^

9! 7! ∗ 2! =^

9 ∗ 8 ∗ 7! 7! ∗ 2! =^

9 ∗ 8 2

Therefore, the number of ways of selecting a pair of 2 different books from 9 books = 36

4. The number of doubles team that can be formed from 5 players = 5C2 = 5! 2! =^

5! (5−2)! 2! =

5∗ 4 ∗ 3! 3! ∗ 2! =^

5 ∗ 4 2 Therefore, the number of doubles team that can be formed = 10

5. As we can observe in this question, order is important here. So this is another case of permutation. So, it is like arranging 7 flags by taking 3 at a time: n = 7, r = 3 nPr = nPr =

𝑛! (𝑛−𝑟)! =^ 7P^3 =^

7! (7−3)! =^

7! 4! =^

7∗ 6 ∗ 5 ∗ 4! 4! =^ 7 * 6 *^ 5 = 210

Thus, the number of unique flags = 210 flags

6. This is obviously a combination question. If the owner of the company and his wife must be members of the 5 person committee, it means that we have already selected 2 out of 5 from 12. Thus, we have 3 members left to be selected from the 10 remaining members. The number of ways to do this = 10C 3 ways =

10! (10−3)! 3! =^

10! 7! ∗ 3! =^

10 ∗ 9 ∗ 8 ∗ 7! 7! ∗ 3! =

10 ∗ 9 ∗ 8 3 ∗ 2 ∗ 1 Therefore, the number of ways of forming the committee = 120 ways

7. There are 20 teams in the league and we know that each game is played between two teams only.

 The match between team A and team B is the same as the match between team B and team A  Thus, for each match to happen, we must select 2 teams only and in this case, the arrangement will not matter

Can you observe we arrived at the keyword SELECT by carefully analyzing the information

given, and making some meaningful inferences? Now, we only have to find the number of ways of selecting 2 teams from 20 teams and then

multiply by 2 since there are two games (home and away) played between any two teams_._ Hence, total number of games played = 2! * 20C 2

= 2 * 20! ( 20 − 2 )! ∗ 2!

= 2 *

20! 18! ∗ 2!

= 2 *

20 ∗ 19 ∗ 18! 18! ∗ 2!

= 2 *

20 ∗ 19 2 = 20 * 19 Thus, the total number of games played = 380 games

REFERENCES

1. Socratic.org/questions
2. E-GMAT questions

APPENDIX

Common Factorials (!)

Take note of the following factorials:

n! = (n – 1)(n – 2)(n – 3)……….2 * 1

0! = 1

1! = 1

𝑛! 𝑛! = 1