Permutations and Combinations
This topic looks at the number of different ways of choosing and arranging objects.
Arranging Objects: Permutations
Example 1
In how many ways can the letters A, B and C be arranged to form three-letter ”words”?
In this case we can list all the possibilities systematically and see that there are six arrangements.
ABC, ACB, BCA, BAC, CBA, CAB
Example 2
Four athletes compete in the 100m final. In how many different orders can they finish (assuming
there are no dead heats)?
Instead of listing all the possibilities we can think of the problem like this.
The number of runners who could cross the line first is 4.
Once the first runner is across the line, there are 3 runners who could be second.
Once the first two runners are across the line, there are 2 runners who could be third.
Once the first three runners are across the line there is just one runner who is fourth.
So the number of different orders for the runners to finish in is:
4 × 3 × 2 × 1 = 24
Factorials
We introduce factorial notation to represent this. 4! = 4 × 3 × 2 × 1 = 24
7! = 7 × 6 × 5 × 4 × 3 × 2 × 1 = 5040
In general
𝑛! = 𝑛(𝑛 − 1)(𝑛 − 2) ⋅ ⋅ ⋅ 3 ⋅ 2 ⋅ 1
We define 0! = 1 (which will be useful later).
General Result
There are 𝑛! different arrangements (or permutations) of 𝑛 distinct objects.
Example 3
How many different four-digit numbers can be made from the set 1, 2, 2, 3?
In this case the objects we have to permute are not all distinct. If we label the two 2s as 2A
and 2B, and start to list the possibilities, we can get an idea of what is happening.
1
1
1
1
2A
2B
3
3
2B 3
2A 3
2A 2B
2B 2A
In fact so far we have only the two distinct four-digit numbers 1223 and 1322.
Overall we will have exactly half of the permutations we would expect ( 4! = 24) if the digits
were all distinct, so there will be twelve permutations.
Example 4
How many different 5-digit numbers can be made from the set 1, 2, 2, 2, 3?
Again by starting to list the possibilities, and distinguishing the 2s, we can see what will happen.
1
1
1
1
1
1
2A
2A
2B
2B
2C
2C
2B
2C
2C
2A
2B
2A
2C
2B
2A
2C
2A
2B
3
3
3
3
3
3
This would just give us one number 12223. We need to divide the number of permutations we
would expect for distinct digits by 6 = 3!. So the total number of permutations of 1, 2, 2, 2, 3
5!
= 20
will be
3!
General result
𝑛!
The number of permutations of 𝑛 objects, of which 𝑟 are identical and the rest distinct is .
𝑟!
We can extend this idea. For instance, the number of permutations of the letters of
11!
= 4989600
MATHEMATICS is
2!2!2!
Example 5
How many different 3-digit numbers can be made from the set of digits
1, 2, 3, 4, 5?
We can think of the problem like this. We have 5 choices for the first digit, 4 for the second
digit and then 3 for the third digit.
So the number of different 3-digit numbers is 5 × 4 × 3 = 60.
5!
Note that this is .
2!
General result
The number of permutations of a group of 𝑟 objects, taken from a group of 𝑛 distinct objects
is denoted by
𝑛!
𝑛
𝑃𝑟 =
(𝑛 − 𝑟)!
Note that to calculate 𝑛 𝑃𝑟 we start with 𝑛 and then multiply together 𝑟 numbers.
𝑛
𝑃𝑟 = 𝑛(𝑛 − 1)(𝑛 − 2)
⋅⋅⋅
(𝑛 − 𝑟 + 1)
Choosing Objects: Combinations
Now we look at the number of ways of choosing objects from a group, when the order does not
matter.
Example 6
In how many ways can you choose two letters from the list A, B, C, D, E?
By listing the possibilities systematically we see that there are 10 different choices.
AB, AC, AD, AE, BC, BD, BE, CD, CE, DE.
How many ways are there of choosing three letters from the list?
Again we can list the possibilities and see that we have 10 different choices.
ABC, ABD, ABE, ACD, ACE, ADE, BCD, BCE, BDE, CDE
Note that these two answers are the same. This is not a coincidence. Choosing three objects
from a group of five is equivalent to leaving two behind.
5!
5!
= 20 and that 5 𝑃3 =
= 60. In the first case we needed to divide
3!
2!
by 2 (= 2!) to get the number of combinations, and in the second case by 6 (= 3!).
Notice also that 5 𝑃2 =
So the number of ways of choosing two objects from five is given by 5 𝐶2 =
same as 5 𝐶3 .
5!
, which is the
3!2!
General result
The number of ways of choosing 𝑟 objects from a group of 𝑛 distinct objects is given by:
( )
𝑛
𝑛!
𝑛
𝐶𝑟 =
=
𝑟!(𝑛 − 𝑟)!
𝑟
Notice that (𝑛 − 𝑟) + 𝑟 = 𝑛 and that, by symmetry 𝑛 𝐶𝑟 =𝑛 𝐶𝑛−𝑟
Example (7 )
Calculate 50
.
3
( )
50
50!
=
3
(47!)(3!)
50 ⋅ 49 ⋅ 48 ⋅ 47 ⋅ ⋅ ⋅ 3 ⋅ 2 ⋅ 1
=
(47 ⋅ 46 ⋅ 45 ⋅ ⋅ ⋅ 3 ⋅ 2 ⋅ 1)(3 ⋅ 2 ⋅ 1)
50 ⋅ 49 ⋅ 48
=
3⋅2⋅1
= 50 ⋅ 49 ⋅ 8
= 19600
Calculators
Your calculator may have 𝑛!, 𝑛 𝑃𝑟 and 𝑛 𝐶𝑟 buttons. Make sure you understand how they work.
Note that it is often quicker (and sometimes more accurate) to use cancelling as in Example 7.