Permutations and combinations
In everyday life, situations arise when we are required to place a number of objects in
a definite order or to select a number of items from a group of objects.
Example 1: We may need to know the number of ways we can hang a number of
pictures on a wall.
Example 2: It may be of interest to know in how many ways we can pick five
integers, from 1 to 90, to play in the Lottery.
These type of problems fall in two categories:
Permutations: The order of arrangement of the items is important
Combinations: The order in which a selection is made is irrelevant.
Example
In how many different arrangements can three paintings of Rembrandt, Picasso and
Van Gogh be hung on a wall?
Let me suppose that:
R=paintings by Rembrandt
P= paintings by Picasso
V= paintings by Van Gogh.
The possible arrangements are:
RPV RVP PRV PVR VPR VRP.
There are three ways to place the first painting, two ways to place the second painting
and 1 way to place the third painting giving us: 3x2x1=6.
Formula for permutations
Unless the number of items is small, as in the previuos example, it’s difficult to list all
the possible arrangements of the given number of objects among themselves.
The number of Permutations of n unlike objects is n!
n! = n ⋅(n − 1)⋅(n − 2)⋅.....⋅ 2 ⋅1,n ≥ 2
If n = 0 0! = 1
Exercise:
Find the number of permutations on 4 letters taken from the name Lisa if each letter is
chosen only once.
Solution:
P=4!=24.
Exercise1: In how many ways can a team of eleven football players be arranged in a
line?
Exercise2: In how many ways can nine books be arranged on a shelf?
1 Permutations of k objects chosen from a set of n objects (k<n)
Let’s suppose now we want to select a number of items k from the total number of
objects n.
Example: We have 5 paintings and we want to place just three of them in a row.
The paintings are A, B, C, D, E.
We have five different ways to place the first picture, four different ways to place the
second picture and three different ways to place the third picture.
So, the number of arrangements we can have is 5x4x3=60.
Generalizing this result, the number of ways in which to arrange k objects chosen
from a group of n objects is given by:
𝑛!
!
𝑃! =
𝑛−𝑘 !
This formula can also be written:
!
𝑃! = 𝑛 ∙ 𝑛 − 1 ∙ 𝑛 − 2 … … … … .∙ (𝑛 − 𝑘 + 1)
Remember that this formula can be used in case of no repetitions (different items).
Example 1: Find the number of permutations on 4 letters taken from the name Sandro
if each letter is chosen only one (no repetition is allowed).
Applying the formula !𝑃! =
!!
!!
= !! =
!!! !
!∙!∙!∙!∙!∙!
!∙!
= 360
Example 2: There are twenty participants in a competition. In how many ways can the
first three positions be filled?
Example 3: If a shelf can hold only 6 books out of a set of nine books, in how many
ways can the six books be arranged?
Example 4: Find the number of possible arrangements of 4 letters from the alphabet if
no repetitions are allowed .
Example 5: Find the number of permutations on 4 different digits from the set:
0,1,2,3,4,5,6,7,8,9
Example 6: Seven people get into a train but only four seats are available. How many
different ways are there of seating four people at a time?
Permutations with restrictions
There are some situations where some restrictions limit the number of possible
arrangements (the position of an item cannot change).
There are situations where some items could be repeated and others cannot.
Example 1: how many different numbers can be formed using the digits 1,2,3 and 4 if:
a) the numbers must all be greater than 4000 and the digits can be used only once.
2 The first position must be 4. There is one way to choose the first digit, 3 ways to
choose the second digit, 2 ways to choose the third digit, 1 way to choose the last
digit. So, the computation will be 1x3x2x1=6. We have just to compute the number of
permutations of 3 items given 3 places (3!).
Example 2: In how many ways can we arrange the letters of the words TWO and
TOO?
Example 3: Find the number of possible arrangements of the word MALTA.
Permutations with repetition
In general, if we have n objects of which p are the same, q are the same and r are the
same, etc., the number of permutations is given by:
𝑃! (!,!,!) =
𝑛!
𝑝! 𝑞! 𝑟!
Example 4: In how many ways can the letters of the word ASSESSMENT be
arranged?
----------------------------------In situations where we are required to find the number of possible arrangements of r
items from a set of n items with repetitions being allowed then, the number of
possible r-items sequences is 𝑛! .
Example: The plates of Italian cars start with a couple of letters from English alphabet
(they can be repeated). How many initials can we get?
In the English alphabet we have 26 letters. We have just to places for them. So, the
result is 𝑛! = 26! = 676.
Glossary
Permutations= Permutazioni
Combinations= Combinazioni
Digit=Cifra
Items=oggetti
Factorial= Fattoriale
Arrangements= Disposizioni
To pick: estrarre
Times= volte oppure “per” (simbolo moltiplicazione)
To hang on=appendere
Shelf=scaffale
Unlike=diversi
Set=insieme
Hold out= tenere, resistere
Plate=targa (auto)
3 Solve in pairs the following problems about permutations:
1. In how many ways can nine books be arranged on a shelf? If the shelf can
hold only six out of the nine books, in how many ways can the six books be
arranged?
2. There are twenty participants in a competition. In how many way can the first
three positions be filled?
3. How many three-digit numbers can be made from the set {1,2,3,4,5} if:
a) the three digit numbers are all different;
b) the three digit numbers are the same;
c) the three digit numbers is greater than 300 (repetition is allowed);
d) all the three digit numbers are the same but the number is odd;
e) all the three digit numbers are different but the number is even.
4. Ten balls are to be placed in a row. Five of the balls are blue, three are red and
two are yellow. In how many ways can the balls be displayed if the balls of the
same colour are indistinguishable?
5. How many football forecasts of ten football matches can be made?
4 Combinations A combination is a selection of items from a given set without any particular order. Example: In how many way can we select two paintings from a set of three paintings? Let’s call A, B and C the three paintings. The possible combinations are AB, AC or BC. Formula for computing combinations: !
𝑃!
𝑛!
!
𝐶! =
=
𝑟!
𝑛 − 𝑟 ! 𝑟!
Therefore, number of combinations is given by the number of permutations of r items chosen from a set of n items over the number of permutations of the r items among themselves. Take note that the n objects are all different! 𝑛
Another way of writing !𝐶! is which is read as “n choose r”. 𝑟
Example 1: In how many ways can a team of 9 players be chosen from a squad of 22 men. !!
𝑃!
22!
!!
𝐶! =
=
= 497420 9!
22 − 9 ! 9!
The following property is valid: !
𝐶! = !𝐶!!! Let’s make an example: 8!
8∙7∙6∙5∙4∙3∙2∙1
8
=
=
= 56
3
3! 5! 3 ∙ 2 ∙ 1 ∙ 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1
8!
8∙7∙6∙5∙4∙3∙2∙1
8
=
=
= 56
5
5! 3! 3 ∙ 2 ∙ 1 ∙ 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1
Example 2: A box of chocolates contains ten different varieties. In how many ways
can four chocolates be chosen if all four are different?
!"
𝐶! =
10!
= 210
6! 4!
Let’s observe that when we talk about permutations we talk about the number of
arrangements, when we talk about combinations we talk about selections.
5 Exercises-simple combinations
1. I want to interview 4 people chosen from a group of 10 people. In how many
ways can I form the sample of 4 people?
2. In how many ways can I extract four cards from a deck of Sicilian cards?
3. There are 6 people who take part in a meeting. They mutually shake hands.
How many handshakes can you count?
4. How many three winning numbers can you get with the ninety numbers of the
lottery?
Combinations with repetition
Let’s introduce this topic with an example.
We want to know the number of combinations we get when flipping a coin four times.
The possible outcomes are Heads or Tails. The order does not matter.
Working empirically, the combinations are:
HHHH
HHHT
HHCC
TTTH
TTTT.
In this example we note that each outcome can be repeated no more than 4-times.
Then, we can apply the formula:
𝐶′!,!
=
𝑛+𝑟−1 !
𝑛+𝑟−1
=
𝑟
𝑟! 𝑛 − 1 !
Let’s apply this formula in our example.
r=4=number of tosses –max number of repetitions
n=2=number of items (outcomes-heads or tails)
𝐶 ! !,!
=
2+4−1 ! 5∙4∙3∙2∙1
2+4−1
=
=
=5
4
4! 2 − 1 !
4∙3∙2∙1
Let’s consider a different example.
In a ice cream parlour there are 5 flavours of ice-cream: banana, chocolate, vanilla,
strawberry and lemon. We can have three scoops and the flavours can be repeated.
How many variations will there be?
In this example n=5 and r=3.
𝐶′!,!
=
5+3−1 ! 7∙6∙5
5+3−1
=
=
= 35
3
3! 5 − 1 !
3∙2∙1
Finally, here is the last example.
In how many ways can we arrange 6 identical objects in 4 boxes (each box can
contain from 0 to 6 objects)?
If we call a, b, c and d the four boxes, we can list the following possibilities:
aaaaaa (all objects in box a)
6 bbbbbb (all objects in box b)
aaaaab (5 objects in the box a and 1 in the box b)
aabbcc (2 objects in a, 2 in b, 2 in c)
and so on….
Applying the formula we observe that n=4 (number of boxes), r=6 (number of
objects-max number of repetitions)
𝐶′!,!
=
9∙8∙7
4+6−1
9
9
=
=
=
= 84
6
6
3
3∙2∙1
Exercises
1. In how many different ways can we assign 8 bars of chocolate to 5 children,
knowing that some children can have more than 1 bar.
2. In how many ways can we place six equal balls in four different boxes?
3. How many combinations can we get when flipping a coin 6 times?
7