Permutations and Combinations
Permutations and combinations are counting tools. They
have vast applications in probability, especially in determining
the number of successful outcomes and the number of total outcomes
in a given scenario. Questions about permutations and combinations
on the Math IIC will not be complex, nor will they require advanced
math. But you will need to understand how they work and how to work
with them. Important to both of these undertakings is a familiarity
The factorial of a number, n!, is the
product of the natural numbers up to and including n:
If you are ever asked to find the number of ways that
elements of a group can be ordered, you simply
need to calculate n
!. For example, if you are asked
how many different ways 6 people can sit at a table with six chairs,
you could either list all of the possible seating arrangements or
just answer 6! = 6
1 = 720.
A permutation is an ordering of elements. For example,
say you’re running for student council. There are four different
offices to be filled—president, vice president, secretary, and treasurer—and
there are four candidates running. Assuming the candidates don’t
care which office they’re elected to, how many different ways can
the student council be composed?
The answer is 4! because there are 4 students running
for office and thus 4 elements in the set.
Say that due to budgetary costs, there are now only the
three offices of president, vice president, and treasurer to be
filled, but now six candidates are running. To handle this situation,
we will now have to slightly change our method of calculating the
number of permutations.
In general, the permutation, nPr,
is the number of subgroups of size r that can be
taken from a set with n elements:
For our example, we need to find 6P3:
Consider the following problem:
a dog show, three awards are given: best in show, first runner-up,
and second runner-up. A group of 10 dogs are competing in the competition.
In how many ways can the awards be distributed?
This problem is a permutation, since the question asks
us to order the top three finishers among 10 contestants in a dog
show. There is more than one way that the same three dogs could
get first place, second place, and third place, and each arrangement
is a different outcome. So, the answer is 10P3 =
10!/(10 – 3)! =
Permutations and Calculators
Graphing calculators and most scientific calculators have
a permutation function, labeled nPr
In most cases, you must enter n
, then press the
button for permutation, and then enter r
will calculate a permutation for you, but often if n
a large number, the calculator can’t calculate n
If this happens to you, don’t give up! In cases like this, your
knowledge of how the permutation functions will save you. Since
you know that 100P3
you can simplify it to 100!
, or 100
98 = 970,200.
A combination is an unordered grouping of a set. An example
of a scenario in which order doesn’t matter is a hand of cards:
a king, ace, and five is the same as an ace, five, and king.
Combinations are represented as nCr
, where unordered subgroups of size r
from a set of size n
. Because the order of the
elements in a given subgroup doesn’t matter, this means that
will be less than nPr
Any one combination can be turned into more than one permutation. nCr
calculated as follows:
Here’s an example:
that a committee of 10 people must elect three leaders, whose duties
are all the same. In how many ways can this be done?
In this example, the order in which the leaders
are assigned to positions doesn’t matter—the leaders aren’t
distinguished from each other in any way, as they were in the student
council example. This distinction means that the question can be
answered with a combination rather than a permutation. We are looking
for how many different groups of three can be taken from a group
There are only 120 different ways to elect 3 leaders,
as opposed to 720 ways when their roles were differentiated.
Combinations and Calculators
There should be a combination function on your graphing
or scientific calculator labeled nCr.
Use it the same way as you use the permutation key.