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Permutations and
Combinations
Fundamental Counting Principle
If there are r ways of performing one operation, s ways of
performing a second operation, t ways of performing a third
operation, and so on…
then the number of ways of performing the operations in
succession are:
r  s  t  ...
Three different mathematics books and five other different books are
to be arranged on a bookshelf. Find out the number of possible
arrangements if the three mathematics books must be kept together.
Answer: 4320 ways
Circular Arrangements
If we are looking for the number of arrangements of items around a
circular table, we have to consider that some arrangements may be
exactly alike.
This arrangement:
This arrangement:
A
B
A
E
C
E
D
is the same as:
B
D
C
Therefore, to solve these circular arrangements, you fix one of the
elements and arrange the remaining items around it. The number of
arrangements of n unlike things in a circle will be:
(n  1)!
Therefore, to solve these circular arrangements, you fix one of the
elements and arrange the remaining items around it. The number of
arrangements of n unlike things in a circle will be:
(n  1)!
If clockwise and anti-clockwise arrangements are considered
identical, then the number of arrangements would be:
1
 (n  1)!
2
Four men are to be seated at a circular table. In how many ways can
this be done?
Answer: 6 ways
Nine beads, all of different colors, are to be arranged on a circular
wire. Two arrangements are not different if they are the same when
the ring is turned over. How many different arrangements are
possible?
Answer: 20160 ways
The letters of the word TUESDAY are arranged in a line, each
arrangement ending with the letter S and starting with the letter D.
In how many ways is this possible?
Answer: 120 ways
How many numbers greater than 40000 can be formed using the
digits 2, 3, 4, 5 and 6 if each digit is used only once in each
number?
Answer: 72 ways
Mutually Exclusive Situations
Two situations are mutually exclusive if situation A occurs, then
situation B cannot occur. Likewise, if situation B occurs, then
situation A cannot occur.
Therefore:
 Number of Permutations   Number of Permutations in   Total Number 




in
which
event
A
occurs
which
A
does
not
occur
of
Permutations

 
 

or
 Number of Permutations in   Total Number   Number of Permutations 


 

which
A
does
not
occur
of
Permutations
in
which
event
A
occurs

 
 

Sheila, Ann and Harvey are fifth year students. Sarah, Jeff and John
are fourth year students and Alan, Chris and Mark are third year
students. In how many ways can these pupils be arranged in a line if
the pupils from each year are kept together?
Answer: 1296
In how many ways can six boys and two girls be arranged in
a line if the two girls must not sit together?
Answer: 30240
If we want to choose r items from a group of n objects, we would
first have n choices, then (n – 1) choices, then (n – 2) choices and
so on, until the r th choice.
This can be solved as:
n(n  1)(n  2)...(n  r  1)(n  r )(n  r  1)...3  2 1
(n  r )(n  r  1)...3  2 1
This is normally called a permutation and is usually written as:
n
n!
P 
r
(n  r )!
Arrangements of Like and Unlike Things
Suppose the following letters are arranged in a row:
a1 , a2 , a3 , b1 , b2
We already know that there would be 5! possible arrangements.
Now suppose we cannot distinguish between each like letter.
So, we want to arrange the letters:
a , a , a , b, b
There will be
5!
3!2!
possible arrangements.
Therefore, the number of arrangements of n things with p of
one kind, q of another kind, r of another and so on, will be:
n!
p !q !r !
In how many ways can 4 red, 3 yellow and 2 green disks be
arranged in a row, if disks of the same color are indistinguishable?
Answer: 1260 ways
Find the number of three-letter arrangements that can be made from the
letters of the word PYTHAGORAS.
Answer: 528
How many different three-digit even numbers can be formed from
the figures 2, 5, 7 and 9 if repetitions are not allowed.
Answer: 6
In how many ways can the letters of the word PARALLEL be arranged
in a row?
Answer: 3360
In how many ways can the letters of the word BANANA be arranged
with the N’s separated?
Answer: 40
Find the number of two-letter arrangements made from the
letters A, B, C and D.
We can see that there are 12 arrangements:
AB, AC, AD, BA, BC, BD, CA, CB, CD, DA, DB, DC
But, notice that many of these arrangements are the same:
AB = BA, AC = CA, AD = DA, BC = CB, BD = DB, DC = CD
So, the number of combinations of two-letters is
one-half of 12, or 6
We know that an arrangement, or permutation, of n things
taken r at a time is defined as:
n!
P
r
(n  r )!
n
But, each selection of r items will have r! items that are the
same. We must remove these r! identical items.
Therefore, the number of combinations of n objects taken r at a
time is:
n n
n!
   Cr 
(n  r )!r !
r
Use a permutation when the order of an arrangement is
important. Use a combination when the order is NOT important.
Find the number of selections of 4 letters that can be made from
the letters of the word SPHERICAL.
Answer: 126
How many of these selections do not contain a vowel?
Answer: 15
How many different committees, each consisting of 3 boys and 2
girls, can be chosen from 7 boys and 5 girls?
Answer: 350
A team of 7 players is to be chosen from a group of 12 players. One of
the 7 is then to be elected as captain and another as vice-captain. In
how many ways can this be done?
Answer: 33264
A group consists of 4 boys and 7 girls. In how many ways can a team of
five be selected if it is to contain at least one member of each sex.
Answer: 441
In how many ways can a team of five be selected from 4 boys and 7
girls if the team is to contain at least 3 boys.
Answer: 91
To find the number of selections of any size from a group, you have to
consider every possibility.
Example:
How many different selections can be made from the five letters a, b, c, d, e?
First Method:
Second Method:
5
C 5
1
5
C  10
2
5
C  10
Total is 31
3
5
C 5
4
5
C 1
5
There are two choices for the letter
a. Either it is included or not.
Likewise, there are two choices for
b and also for c, d, and e. So,
5
there are: 2  2  2  2  2  2
choices. But, we must remove the
case where no letters are included.
So, there are: 5
2  1  31
Divisions into Groups
Example:
The letters a, b, c, d, e, f, g, h and I are to be divided into three
groups containing 2, 3 and 4 letters respectively. In how many ways
can this be done?
Answer:
9!
9!
7!

1 

1260
7!2! 4!3!
2!3!4!
In general, the number of ways of dividing (p + q + r) unlike things into
three groups containing p, q and r things respectively is:
( p  q  r )!
p !q !r !
Find the number of ways that 12 people can be arranged into groups if
there are to be:
a) Two groups of 6 people.
Answer: 462 ways
b) 3 groups of four people.
Answer: 5775 ways
How many different whole numbers are factors of the number below?
2  3 5  7 1113
Answer: 64
How many different selections can be made from the letters of the
word INABILITY?
Answer: 255
How many subsets of a set of 10 elements have either
3 or 4 elements?
Answer: 330
To participate in a state lottery, a person must choose 4
numbers from 1 to 40. In how many ways can this be
done if:
a) the order of the choice matters
b) the order of the choice does NOT matter.
Answer: a) 2193360
b) 91390
There are 10 seats in a row in a waiting room. There are six people in
the room.
(a) In how many different ways can they be seated?
(b) In the group of six people, there are three sisters who must sit
next to each other. In how many different ways can the group be
seated?
Answer:
(a) 151200
(b) 10080
M06/HL1/19