


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 IC 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
with factorials.
Factorials
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
the n 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 (which
would take a while) or just answer 6! = 6 _{ }5 _{ }4 _{ }3 _{ }2 _{ }1 =
720.
Permutations
A permutation is an ordering of elements.
For example, say you’re running for student council. There are six
different offices to be filled—president, vice president, secretary, treasurer,
spirit coordinator, and parliamentarian—and there are six 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 6! because there are 6 students running
for office, and thus, 6 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. The same 6 candidates are still running. To handle this situation,
we will now have to change our method of calculating the number
of permutations.
In general, the permutation, nP_{r},
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 _{6}P_{3}:
Consider the following problem:

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 _{10}P_{3} = ^{10!}⁄_{(10
3)!} = ^{10!}⁄_{7!} =
720.
Permutations and Calculators
Graphing calculators and most scientific calculators have
a permutation function, labeled nP_{r}.
In most cases, you must enter n, then press the
button for permutation, and then enter r. This
will calculate a permutation for you, but if n is
a large number, the calculator often cannot calculate n!.
If this happens to you, don’t give up! In cases like this, your
knowledge of the permutation function will save you. Since you know
that _{100}P_{3} is ^{100!}⁄_{(100
3)!} you can simplify it to ^{100!}
/_{97!} , or 100 99 98 = 970,200.
Combinations
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, an ace, and a five is the same as an ace,
a five, and a king.
Combinations are represented as _{n}C_{r} ,
or , where unordered subgroups of size r are selected
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 _{n}P_{r}.
Any one combination can be turned into more than one permutation. _{n}C_{r} is
calculated as follows:
Here’s an example:

In this example, the order in which the leaders are assigned
to positions doesn’t matter—the leaders aren’t distinguished from
one another in any way, unlike 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 of 10:
There are only 120 different ways to elect three 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 _{n}C_{r}.
Use it the same way you use the permutation key.
