- Permutations

PERMUTATIONS

The different arrangements of a set of objects are called permutations.
LESSON 1
Determine the number of permutations of the letters A, B, C and D.
SOLUTION
We have:
4 options for the first letter.
3 options for the second letter
2 options for the third letter
1 option for the fourth letter.
Number of permutations is \(4×3×2×1=24\)
This is equivalent to \(4!\)
The number of permutations of \(n\) distinct object is \(n!\)

LESSON 2
The digits 1, 2, 3, 4 and 5 are arranged in random order, to form a five – digit number. How many different five – digit numbers can be formed?
SOLUTION
Number of five – digit numbers is \(5!=120\)

LESSON 3
A test consists of 4 algebra questions A, B, C and D and 4 trigonometry questions G, H, I and J. The examiner plans to arrange all 8 questions in a random order, regardless of topic. How many different arrangements are possible?
SOLUTION
Number of arrangements is \(8!=40320\)

PERMUTATIONS OF \(r\) OUT OF \(n\) DISTINCT OBJECTS
The number of permutations of \(r\) out of \(n\) distinct objects is \(\displaystyle ^nP_r ={n!\over (n-r)!}\)
LESSON 1
Determine the number of arrangements of 5 letters of the word COMPUTER.
SOLUTION
There are
8 options for the first letter
7 options for the second letter
6 options for the third letter
5 options for the fourth letter
4 options for the fifth letter
The number of arrangements is
\(8×7×6×5×4=6720\)
This is equivalent to
\(\displaystyle {8!\over (8-5)!}=6720\)

LESSON 2
A bag contains 9 discs numbered 1, 2, 3, 4, 5, 6, 7, 8, 9. Amya chooses 4 discs at random, without replacement, and places them in a row. How many different 4 digit numbers can be made?
SOLUTION
There are
9 options for the first digit
8 options for the second digit
7 options for the third digit
6 options for the fourth digit
The number of 4-digit numbers is
\(9×8×7×6=3024\)
This is equivalent to
\(\displaystyle {9!\over (9-4)!}=3024\)

PERMUTATIONS WITH REPEATED OBJECTS
Number of arrangements of \(n\) objects given repeated objects
\(\displaystyle {(number \:of \:objects)!\over (number\: of \:repeated \:objects)!}\)
LESSON 1
The five letters of the word NEVER are arranged in random order in a straight line. How many different arrangements of the letters are possible?
SOLUTION
Since there are 2 E’s these are NOT distinct, therefore when their positions are switched the arrangement is the same.
Number of arrangements is
\[{5!\over 2!}=60\]

LESSON 2
Each of the 7 letters in the word DIVIDED is printed on a separate card. The cards are arranged in a row. How many different arrangements of the letters are possible?
SOLUTION
There are 7 letters with 2 Ds and 2 Is
Number of arrangements\(\displaystyle ={7!\over 2!3!}=420\)

PERMUTATIONS WITH CONDITIONS
LESSON 1
5 of the 7 letters A, B, C, D, E, F, G are arranged in a random order is a straight line. How many arrangements end with a vowel?
SOLUTION
We have
2 options for the last letter (A or E)
6 options for the first letter
5 options for the second letter
4 options for the third letter
3 options for the fourth letter
Number of arrangements is
\[2×6×5×4×3=720\]

LESSON 2
Find how many numbers between 5000 and 6000 can be formed using the digits 1, 2, 3, 4, 5, 6.
SOLUTION
We have
1 option for the first digit since it has to be 5
5 options for the second digit
4 options for the third digit
3 options for the fourth digit
Number of numbers is \(1×5×4×3=60\)

LESSON 3
Daveed has 11 different CDs, of which 6 are pop music, 3 are jazz and 2 are classical.
How many different arrangements of all 11 CDs on a shelf are there if the jazz CDs are all next to each other?
SOLUTION
There are 11 CDs but we treat the 3 Jazz CDs as 1 object which can be arranged in 3! ways within themselves. We therefore now have 9 objects to arrange.
Number of arrangements is \(3!×9!=2,177,280\)

LESSON 4
The back row of a cinema has 12 seats, all of which are empty. A group of 8 people, including Mary and Frances, sit in this row. Find the number of different ways they can sit in these 12 seats if
(i) there are no restrictions,
(ii) Mary and Frances do not sit in seats which are next to each other,
(iii) all 8 people sit together with no empty seats between them.
SOLUTION
(i) The first person has 12 seats to choose from
The second has 11
The third has 10 and so on

The eight has 5 seats to choose from
\(^{12}P_8=19,958,400\)
(ii) When they sit together
\[2!×^{11}P_7=3,326,400\]
When they do not sit together \(=19,958,400-3,326,400=16,632,000\)
(iii) Group the 8 persons together and they can be arranged in 8! ways and can be placed in 5 positions.
\[5×8!=201,600\]

LESSON 5
The digits of the number 1 244 687 can be rearranged to give many different 7 – digit numbers. How many of these 7 – digit numbers are even?
SOLUTION
We have two scenarios to consider: (1) the number ends with 4 and (2) the number ends with 2, 6 or 8.
If the number ends with a 4 we have
1 option for the last digit since the 4s are repeated
We have 6 distinct objects which can be arranged in 6! ways.
Number of arrangements is \(1×6!=720\)
If the number ends with a 2, 6 or 8 we have
3 options for the last digit
We have 6 objects with 2 repeated objects which can be arranged in 6!/2! ways
Number of arrangements is \(\displaystyle 3×{6!\over 2!}=1,080\)
Total number of arrangements is \(720+1080=1800\)

LESSON 6
Seven friends together with their respective partners all meet up for a meal. To commemorate the occasion, they arrange for a photograph to be taken of all 14 of them standing in a line.
(i) How many different arrangements are there if each friend is standing next to his or her partner?
(ii) How many different arrangements are there if the 7 friends all stand together and the 7 partners all stand together?
SOLUTION
(i) Each of 7 couples can be arranged in \(7!\) ways and each couple can be arranged in \(2!\) ways
Number of arrangements \(=7!×2^7=645,120\)
(ii) The two groups can be arranged in 2! ways and within each group the friends can be arranged in 7! ways.
Number of arrangements is \(2!×7!×7!=50,803,200\)

PERMUTATIONS AROUND A CIRCLE
\(n\) objects can be arranged in a circle \((n-1)!\) ways
LESSON 1
At a dinner party 5 men sit at a round table. In how many ways can they sit?
SOLUTION
\((5-1)!=24\) ways

LESSON 2
In a meeting 6 men and 6 women sit around a round table. Determine the number of seating arrangements if,
(i) there are no restrictions,
(ii) the men and women alternate.
SOLUTION
(i) \((12-1)!=39,916,800\)
(ii) \((6-1)!×6!=86,400\)