Permutation and Combination
In this permutation we will learn about
Permutations
A permutation is an arrangement of a set of objects
for which the order of the objects is important.
Show the number of ways of arranging the letters of the word CAT:
CAT
CTA There are six arrangements or permutations
ACT of the word CAT. By changing the order
ATC of the letters, you have a different
TCA permutation.
TAC
Using the Fundamental Counting Principle, we would have
3 x 2 x 1 number of distinct arrangements or permutations.
In calculating permutations,
we often encounter expressions such
as 3 x 2 x 1. The product of consecutive natural numbers in
decreasing order down to the number one can be
Represented using factorial notation: 3 x 2 x 1 = 3!
Finding the Number of Permutations
1. Determine the number of ways that seven boys can line up.
7 x 6 x 5 x 4 x 3 x 2 x 1 = 5040
There are 5040 different ways that seven boys can line up.
The number of permutations of n
different objects taken all at a time is
n!
2. In how many ways can the letters of the word HARMONY
be arranged?
7! = 5040
There are 5040 ways of arranging the seven letters.
Finding the Number of Permutations
How many three-letter permutations can be formed from the
letters of the word DINOSAUR?
8 x ____
____ 7 x ____
6 There are 336 ways.
1st 2nd 3rd
336 represents the number of permutations
of eight objects taken three at a time.
This may be written as 8P3. (This is read as “eight permute three”.)
In general, if we have n objects but only want to select r
objects at a time, the number of different arrangements is:
nPr = n (n-1) (n-2) ….((n-(r-1))
= n (n-1) (n-2) ….(n-r +1)
n (n-1) (n-2) ….(n-r +1) (n-1)!
(n-1)!
n!
n Pr The number of permutations of n objects taken n
(n r)!
at a time is: nPn = n!
Permutations with Repetition or not Distinct
How many four-letter words can be made using
the letters of PEER?
4P4 = 4!
Note that PEER and PEER are not distinguishable,
therefore, they count as one permutation.
When this happens, you must divide by the factorial of
the number of repeated terms.
For PEER, the number of arrangements is:
4!
12 arran gemen ts
2!
The number of permutations of n objects taken n at a time, if
there are a alike of one kind, and b alike of another kind, c alike
of a third kind, and so on, is:
n!
a!b!c!...
Finding the Number of Permutations
A special case of this formula occurs when n = r.
3!
3P3 = 3! 3 P3
(3 3)!
3!
3 P3
0!
From these two results, we can see that 0! = 1.
To have meaning when r = n, we define 0! as 1.
1. Find the value of each expression:
a) 6P3 = 120 b) 10P7 = 604 800
2. Using the letters of the word PRODUCT, how many
four-letter arrangements can be made?
7P4 = 840 There are 840 arrangements.
Permutations with Restrictions
1. In how many ways can the letters of the word ENGINEERING
be arranged?
11!
number of arrangements
3! 3! 2! 2!
= 277 200
2. Naval signals are made by arranging coloured flags in a vertical
line. How many signals can be made using six flags, if you have:
a) 3 green, 1 red, and 2 blue flags?
b) 2 red, 2 green, and 2 blue flags?
6! 6!
number of arrangements number of arrangements
3! 2! 2! 2! 2!
= 60 = 90
Finding the Number of Permutations
1. How many six-letter words can be formed from the letters
of TRAVEL? (Note that letters cannot be repeated)
a) If any of the six letters can be used:
6P6 = 6!
= 720
b) If the first letter must be “L”:
1 x 5P5 = 1 x 5!
= 120
c) If the second and fourth letters are vowels:
___ ___
2 ___ ___
1 ___ ___
2 x 1 x 4! = 48 V V
d) If the “A” and the “V” must be adjacent:
(Treat the AV as one letter - this grouping can be arranged 2! ways.)
5! x 2! = 240
Finding the Number of Permutations
2. A bookshelf contains five different algebra books and seven
different physics books. How many different ways can these
books be arranged if the algebra books are to be kept together?
For the five algebra books: 5!
The five algebra books are considered as one item, therefore,
you have eight items (7 + 1) to arrange: 8!
Total number of arrangements = 8! x 5!
= 40 320 x 120
= 4 838 400
3. In how many ways can six boys and six girls be arranged on
a bench, if no two people of the same gender can sit together?
b ___
___ g ___
b ___
g ___
b ___
g ___
b ___
g ___
b ___
g ___
b ___
g
___
g ___
b ___
g ___
b ___
g ___
b ___
g ___
b ___
g ___
b ___
g ___
b
Boys: 6! The total number of arrangements is
Girls: 6! 6! x 6! x 2 = 1 036 800.
Finding the Number of Permutations
4. In how many ways can the letters of the word
MATHPOWER be arranged if:
a) there are no further restrictions?
9! = 362 880
b) the first letter must be a P and the last letter an A?
1 x 7! x 1 = 5040
c) the letters MATH must be together?
6! x 4! = 17 280
d) the letters MATH must be together and in that order?
1 x 6! = 720
Finding the Number of Permutations
6. How many arrangements of four letters are there from
the word PREACHING?
9P4 = 3024
7. There are six different flags available for
signal. A signal consists of at least four flags
tied one above the other. How many
different signals can be made?
6P4 + 6P5 + 6P6 = 1800
Permutations with Restrictions
3. Find the number of arrangements of the letters of UTILITIES:
a) if each begins with one I and b) if each begins with
the second letter is not an I. exactly two I’s.
___ 6 ___
3 x___x 7! 3 x___x
___ 2 ___x___
6 6!
3 6 7! 3 2 6 6!
arrangements arrangements
3! 2! 3! 2!
= 7560 = 2160
4. How many arrangements are there, using all the letters of the
word REACH, if the consonants must be in alphabetical order and
together?
Permutation with Repetition Allowed
1. How many numbers can be made from the digits 2, 3, 4, and 5,
if no digit can be repeated?
4! + 4P3 + 4P2 + 4P1 = 64
Example
1. A team of 7 people is to be chosen from 8 men and 9
women.
In how many ways can this be done if there must be 4 men
and 3 women on the team.
Solution
Number of selections:
8C
4 men from 8 different men 4
3 women from 9 different women 9C
3
8C 9C = 5880 teams
4 3
2. A team of 7 people is to be chosen from 8 men and 9
women. In how many ways can this be done if there
must be more men than women on the team.
Solution
M W 8C 9C
7 0 =8
7 0 8C 9C
6 1 = 252
6 1 8C 9C
5 2 = 2016
5 2 8C 9C
4 3 = 5880 (from (a))
4 3
Total = 8 + 252 + 2016 + 5880
= 8156 teams
3. A team of 7 people is to be chosen from 8 men and 9 women. In
how many ways can this be done if there must be 4 men and 3
women and one of the women, Anya, refuses to be on the team
if one particular man, Bertie, is on the team.
Solution
Number of teams with 4 men and 3 women = 5880 (from (a))
Number of teams that have both Anya and Bertie on them:
7C
3 more men from 7 are needed 3
2 more women from 8 are needed 8C
2
7C 8C = 980
3 2
Answer = 5880 – 980
= 4900 teams
4. Three letters are selected at random form the letters of the
word BIOLOGY. Find the total number of selections.
Solution
No letters ‘O’
One letters ‘O’
Two letters ‘O’
Case – 1 Number of selections with no letters ‘O’
Number of ways to choose three letters from B, I, L, G, Y
Case – 2 Number of selections with One letters ‘O’
Number of ways to choose two letters from B, I, L, G, Y
Case – 3 Number of selections with Two letters ‘O’
Number of ways to choose One letters from B, I, L, G, Y
