Permutations and Combinations | Permutations

Permutation – Definition, Formula, Practical Examples

In this article, we look at Permutation – definition, types and examples.

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Permutation is the arrangement of items in order
Given n number of items, r items can be permutated where $r \le n$ .
Permutation can be done with or without repetition.

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Permutation is the ordered arrangement of objects taking some or all at a time.

Permutation of n objects taken r at a time

The number of ordered arrangements of $n$ items taken $r$ at a time is $^nP_r$ .

$^nP_r = \frac{n!}{ (n-r)!}$ where $r \le n, r \in W, n \in N.$

Note that $W$ is a set of Whole numbers, $N$ is a set of Natural numbers.

Example 1:

If 7 athletes compete in an event, in how many ways can gold, silver and bronze medals be awarded ?

Solution

(Using Permutation definition)

This is an example of permutation because the order of arrangement for the awards is important. This is because the order determines who gets gold, silver or bronze. You can’t award bronze to the first or gold to the last.

Total athletes $=7$ , Awards $=3$

Out of 7 athletes, we pick any $3$ at a time and permute or rotate their positions. That’s,

$^nP_r$ = $^7P_3$

$^7P_3 = \frac{7!}{(7-3)!} =\frac{ 7!}{4!}=\frac{7 \times 6 \times5 \times4 \times3 \times2 \times 1}{4\times3\times2\times1}$

$=7\times6\times5$

$= 210$ ways

Alternative Approach

(Using the Multiplication Rule of principle of counting)

We have 7 athletes and we are selecting 3 for three awards. We will select gold winner, followed by silver winner then followed by bronze winner. This is a clear application of multiplication rule of principle of counting.

Let’s look at our different chances during the selection of the individual winners in order.

Gold award : In awarding gold, it can go to any of the 7 athletes. Thus, we have 7 choices.

Silver award: In awarding the silver, we are left with 6 people or choices to select from.

Bronze award : After selecting two people for gold and silver already, we are left with just 5 people or choices to select for bronze.

Therefore, to select gold and silver and bronze winners, we have 7×6×5= 210 ways

Example 2:

In a committee of 15 members, in how many ways can three officers — president, vice president, and treasurer be selected?

Solution

This is permutation because the order in which president, vice president and treasurer are selected matters. For example, selecting person A as president and person B as vice president is different from selecting person B as president and person A as vice.

Total members $=15$ , Number of positions $=3$

Out of 15 members, we pick any 3 at a time for the three positions. That’s,

$^15P_3 = \frac{15!}{(15-3)!}$

$= \frac{15!}{12!}$ $=\frac{15\times14\times13\times12\times11\times10\times9\times8\times7\times6\times5\times4\times3\times2\times1}{12\times11\times10\times9\times8\times7\times6\times5\times4\times3\times2\times1}$

$=15\times14\times13$

$= 2,730$ ways

Alternative Approach

Using the multiplication rule of principle of counting

We have 15 members and we are selecting President, followed by a vice president and then a treasurer.

Selecting President : In selecting President,we can select any of the 15 members. Therefore, we have 15 choices.

Selecting Vice-president: In selecting Vice-president, we have 14 members or choices left to select from.

Selecting Treasurer: After selecting two members already, we are left with just 13 members or choices to select from.

Therefore, to select president , vice and treasurer, we have $15\times14\times13= 2,730$ ways

Example 3:

In how many ways can 4 men lodge at a hotel if 7 rooms are available?

Solution

This is permutation because the order in which the 4 men lodge at the 7 different rooms at the hotel can be different. For example, if Persons A, B, C and D were to lodge at any of rooms 1,2,3,4,5,6, or 7, one possibility is Person A chooses room 1, Person B chooses room 2, Person C chooses room 3, person D chooses room 4. Another possiblity if person A chooses room 7, person B chooses room 3, person C chooses room 2, person D chooses room 5. The possiblities are numerous and order differs in each case.

Total men $=4$ , Number of rooms in the hotel $=7$

Out of 7 rooms, the 4 men select a room to lodge in. Thats, out of 7 rooms, 4 rooms are selected. That’s,

$^7P_4 = \frac{7!}{(7-4)!}$

$= \frac{7!}{3!}$

$=\frac{7\times6\times5\times4\times3\times2\times1}{3\times2\times1}$

$=7\times6\times5\times4$

$= 840$ ways

Alternative Approach

Using the multiplication rule of principle of counting

We have 7 rooms and we are selecting 4 rooms for the four men.

First selection : In selecting a room for the first man, he has 7 choices.

Second seletion: In selecting a room for the second man, he has 6 choices.

Third seletion: In selecting a room for the third man, he has 5 choices.

Therefore, to select four rooms for the 4 men out of the 7 available rooms, we have $7\times6\times5 = 840$ ways

Example 4:

In how many ways can 12 students sit on a bench if only 5 seats are vacant?

Solution

This is permutation because the order in which the 12 students occupy the 5 vacant seats will be different in many ways. For example, from the twelve students, student one could occupy the first vacant seat or the last or any other vacant seat.

Total students $=12$ , Number of vacant seats on bench $=5$

Out of 12 students, only 5 vacant seats are available to occupy.This implies that, we can only select 5 students from the 12 students by minding the order of arranging them on the vacant seats. In order words, we will permutate 5 students out of the 12 available.

That’s,

$^{12}P_5 = \frac{12!}{(12-5)!}$

$= \frac{12!}{7!}$ $=\frac{12\times11\times10\times9\times8\times7\times6\times5\times4\times3\times2\times1}{7\times6\times5\times4\times3\times2\times1}$

$=12\times11\times10\times9\times8$

$= 95,040$ ways

Alternative Approach

Using the multiplication rule of principle of counting

We have 12 students and we are selecting 5 students to fill 5 vacant seats on the bench.

First student : The first student can be selected from the 12 students. That is, we have 12 choices to select a student from to fill first vacant seat.

Second student : The second student from the remaining 11 students can be selected in 11 ways.

Third student : The third student from the remaining 10 students can be selected in 10 different ways to fill the third vacant seat.

Fourth student : The fourth student from the remaining 9 students can be selected in 9 different ways to fill the fourth vacant seat.

Fifth student : The fifth student from the remaining 8 students can be chosen in 8 different ways to fill the last vacant seat.

Therefore, the 5 vacant seats can be occupied by the 12 students in $12\times11\times10times9times18$

$= 95,040$ ways

Understanding the Question

At this point, you may be wondering that we first selected 5 students from 12 students and then arranged them to fill 5 vacant seats so why weren’t these two steps considered separately?

Indeed, it is possible to consider each step separately. From 12 students, we need just 5 students. This is selection or combination. The order isn’t important. That’s,

$^{12}C_5$

$=\frac{12!}{(12-5)!5!}$

$=\frac{12!}{7!5!}$

Now, this gives the number of ways we can select or combine 5 students from 12 students. Order is not important here.

However, to arrange the 5 selected students to occupy the 5 vacant seats, the order is important. This is because student one can occupy space 1 or 2 or any in one possibility and another in other possibility. This is permutation.

To arrange 5 students on 5 vacant seats, we write

$^5P_5$

$=\frac{5!}{(5-5)!}$

$=\frac{5!}{0!}$

$=\frac{5!}{1}$

$=5!$

If we select the 5 students from 12 students and then arrange the 5 students on the 5 vacant seats, we have

$=\frac{12!}{7!5!} \times 5!$

$=\frac{12!}{7!}$

This resolves to the expression for permutation we had earlier above. That’s,

$=\frac{12\times11\times10\times9\times8\times7\times6\times5\times4\times3\times2\times1}{7\times6\times5\times4\times3\times2\times1}$

$= 95,040$ ways

Example 5:

In how many ways can 6 distinct objects be arranged in a row taking 3 objects at a time?

Solution

This is permutation because arrangement of the objects is done in order.

Total distinct objects $=6$ , selection at a time $=3$

Out of 6 distinct objects, 3 are taken and arranged at a time. That’s,

$^6P_3 = \frac{6!}{(6-3)!}$

$= \frac{6!}{3!}$

$=\frac{\times6\times5\times4\times3\times2\times1}{3\times2\times1}$

$=\times6\times5\times4$

$= 120$ ways

Alternative Approach

Using the multiplication rule of principle of counting

We have 6 distinct objects and we are selecting and arranging 3 objects.

Selecting and arranging first item : To select and arrange first item, we can choose from any of the 6 distinct items. That’s, we have 6 different ways.

Selecting and arranging second item : To select and arrange second item, we can choose from any of the 5 remaining distinct items. That’s, we have 5 different ways.

Selecting and arranging second item : To select and arrange third item, we can choose from any of the 4 remaining distinct items. That’s, we have 4 different ways.

Therefore, to select 3 different items from 6 distinct objects, it can be done in $6\times5\times4 = 840$ ways

Permutation of all n distinct objects

The number of ordered arrangements of n items taken n at a time is $^nP_n$ .

$^nP_n = \frac{n!}{ (n-n)!}=\frac{n!}{ 0!}=\frac{n!}{1}=n$ where $n \in N.$

That’s, if you have, say 5, distinct items and you arrange or permutate all the 5 items, it is $5!$ .

Example 1:

In how many ways can the letters P,Q,R be arranged in a row?

Solution

This is permutation because arrangement of the letter is done in order.

Total number of letters $=3$ , number of letters to arrange at a time $=3$

Out of 3 letters, we arrange all 3 at a time.

$^3P_3 = \frac{3!}{(3-3)!}$

$= \frac{3!}{0!}$

$= \frac{3!}{1}$

$= 3!$

$=3\times2\times1$

$= 6$ ways

Alternative Approach

Using the multiplication rule of principle of counting

We have 3 letters and arranging all 3 letters.

Arranging first letter: There are three slots where we can put a first letter.

___ ___ ___

Arranging second letter: There are two slots left where we can put the second letter.

_p__ ___ ___

Arranging third letter: There is one slot left where we can put the last letter.

_p__ _Q__ ___

Therefore, to arrange all the three letters, we have $3\times2\times1 = 6$ ways

NOTE: The order of arrangement of letters can change buy the number of choices at each level will be the same.

Example 2:

In how many ways can the letters of the word TABLE be arranged?

Solution

This is permutation because arrangement of the letters are done in order.

Total number of letters $=5$ , number of letters to arrange at a time $=5$

Out of 5 letters, we arrange all 5 at a time.

$^5P_5= \frac{5!}{(5-5)!}$

$= \frac{5!}{0!}$

$= \frac{5!}{1}$

$= 5!$

$=5\times4\times3\times2\times1$

$= 120$ ways

Alternative Approach

Using the multiplication rule of principle of counting

We have 5 letters and arranging all 5 letters.

Arranging first letter: There are 5 slots where we can put a first letter.

___ ___ ___ ___ ___

Arranging second letter: There are 4 slots left where we can put the second letter.

_T__ ___ ___ ___ ___

Arranging third letter: There are 3 slots left where we can put the third letter.

_T__ _A__ ___ ___ ___

Arranging fourth letter: There are 2 slots left where we can put the fourth letter.

_T__ _A__ _B__ ___ ___

Arranging fifth letter: There is 1 slot left where we can put the fifth or last letter.

_T__ _A__ _B__ _L__ ___

Therefore, to arrange all the 5 letters, we have $5\times\times4\times3\times2\times1 = 6$ ways

NOTE: The order of arrangement of letters can change but the number of choices at each level will be the same. For example, we can have

_B__ _L__ _T__ _A__ ___

instead of

_T__ _A__ _B__ _L__ ___

Example 4:

In how many ways can you arrange the letters of the word SOCIAL ?

Solution

This is permutation because arrangement of the letters are done in order.

Total number of letters $=6$ , number of letters to arrange at a time $=6$

Out of 6 letters, we arrange all 6 at a time.

$^6P_6= \frac{6!}{(6-6)!}$

$= \frac{6!}{0!}$

$= \frac{6!}{1}$

$= 6!$

$=6 \times 5\times4\times3\times2\times1$

$= 720$ ways

Alternative Approach

Using the multiplication rule of principle of counting

We have 6 letters and arranging all 5 letters.

Arranging first letter: There are 6 slots where we can put a first letter.

___ ___ ___ ___ ___ ___

Arranging second letter: There are 5 slots left where we can put the second letter.

_S__ ___ ___ ___ ___ ___

Arranging third letter: There are 4 slots left where we can put the third letter.

_S__ _O__ ___ ___ ___ ___

Arranging fourth letter: There are 3 slots left where we can put the fourth letter.

_S__ _O__ _C__ ___ ___ ___

Arranging fifth letter: There is 2 slot left where we can put the fifth letter.

_S__ _O__ _C__ __I_ ___ ___

Arranging sixth letter: There is 1 slot left where we can put the fifth letter.

_S_ _O__ _C__ __I_ _A__ ___

Therefore, to arrange all the 6 letters, we have $6 \times 5\times 4\times 3\times2\times 1 = 6$ ways

NOTE: The order of arrangement of letters can change but the number of choices at each level will be the same. For example, we can have

_S_ _O__ _C__ __I_ _A__ ___

instead of

_A_ _I__ _C__ __S_ _O__ ___

Permutation with Repetition

The above permutations are done without repetition.

The permutation of $n$ different items taken $r$ at a time, when each can be repeated any number of times is $n^r$ .

Proof

Consider $r$ positions to fill with $n$ different items.

For the first position, we have $n$ choices.

For the second position, regardless of what we chose for the first position, we still have $n$ choices.

Similarly, for the third position, regardless of the choices made for the first two positions, we still have $n$ choices, and so on until we fill all $r$ positions.

By the multiplication principle (the fundamental principle of counting), the total number of permutations with repetition is the product of the number of choices at each position, which is $n×n×n×n$ (repeated r times), equal to $n^r$ .

Thus, the formula $n^r$ represents the total number of permutations when choosing $r$ items from $r$ different items with repetition allowed.

Example 1:

In how many ways can the letters P,Q,R be arranged in a row?

Solution

Number of letters $=3$

Number of letters to arrange at a time $=3$

$\implies 3^3$

$=27$ ways

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Permutation – Definition, Formula, Practical Examples

Douglas Tawiah Dwumor

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Permutation of n objects taken r at a time

Permutation of all n distinct objects

Permutation with Repetition

Proof

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