Counting and Probability

Permutations and Combinations

Factorials

A factorial of a natural number represents a product of all the natural numbers less than or equal to the number.
A factorial is the product of all natural numbers less than or equal to the indicated natural number. The symbol for a factorial is an exclamation point.
n!=n(n1)(n2)...1n! = n \cdot (n-1) \cdot (n-2) \cdot . . . \cdot 1
For example:
5!=54321=1205! = 5\cdot 4 \cdot 3 \cdot 2 \cdot 1=120
The expression 0!0! may be a result of evaluating formulas with factorials. Since zero is not a natural number, the formula cannot be used. The factorial of zero is defined as 1: 0!=10! = 1. Factorials are used in counting and probability and are often simplified. To simplify 8!5!!\frac {8!}{5!} !, write it in expanded form and simplify the factors in the numerator and denominator.
8!5!=8765432154321=876=336\begin{aligned}\frac {8!}{5!}&= \frac{8 \cdot 7 \cdot 6 \cdot {\color{#c42126}5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}}{{\color{#c42126}5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}}\\&= 8 \cdot 7 \cdot 6\\&= 336\end{aligned}
Notice that the factors in the numerator and denominator do not have to be written out completely because all the factors in the denominator are also factors in the numerator.
8!5!=8765!5!=876=336\begin{aligned}\frac {8!}{5!}&= \frac{8 \cdot 7 \cdot 6 \cdot {\color{#c42126}5!}}{{\color{#c42126}5!}}\\&= 8 \cdot 7 \cdot 6\\&= 336\end{aligned}

Permutations

A permutation is a selection of elements of a set such that order matters; the number of permutations of nn objects chosen rr at a time can be determined by a formula.

A permutation is the selection and arrangement of objects in a certain order. For example, there are many ways five siblings can arrange themselves in a row of five seats at the movie theater. The order in which the siblings are seated is different in each arrangement, so the situation involves permutations. To determine the number of unique ways the siblings can be ordered, use a factorial.

The factorial 5!5! counts the number of ways five objects can be arranged in different orders.
5!=54321=1205! = 5\cdot 4 \cdot 3 \cdot 2 \cdot 1=120
So, there are 120 unique ways the five siblings can be arranged.

When there are more objects to choose from than objects that will be arranged, a formula is used to calculate the number of permutations.

The number of permutations of nn objects chosen rr at a time is:
P(n,r)=n!(nr)!P(n,r) = \frac {n!}{(n-r)!}
Note that each object in a permutation can be chosen only once.
Step-By-Step Example
Using the Permutation Formula
A music fraternity has four different positions available: president, vice president, treasurer, and communications chairperson. They are choosing from a group of 12 students to fill all four positions. How many different ways are there to choose four students from a group of 12 to fill four positions?
Step 1

Determine whether the situation involves permutations. If so, find the values for nn and rr.

The order of the students chosen for the positions matters because only one student can be chosen for each position. For instance, the student chosen for president is different from a student chosen for vice president. The situation involves permutations.

There are n=12n=12 students to choose from. There are r=4r=4 students who will be chosen for the available positions.

Step 2
Substitute values in the permutation formula and simplify.
P(n,r)=n!(nr)!P(12,4)=12!(124)!=12111098!8!=1211109=11,880\begin{aligned}P(n,r) &= \frac {n!}{(n-r)!} \\ P(12,4) &= \frac {12!}{(12-4)!} \\ &= \frac {12 \cdot 11 \cdot 10 \cdot 9 \cdot {\color{#c42126}8!}}{{\color{#c42126}8!}} \\ &= 12 \cdot 11 \cdot 10 \cdot 9 \\ &= 11,880\end{aligned}
Solution
There are 11,880 different ways to choose 4 officeholders from a group of 12 students.

The multiplication principle can also be used to identify the number of permutations in a situation. If there are 12 students possible who might be chosen for four different music fraternity positions, then there are 12 possible choices for the first position. After the first position is chosen, there are 11 possible choices for the second position, 10 possible choices for the third position, and 9 possible choices for the fourth position. Using the multiplication principle, there are 11,880 ways to choose students to fill four positions.
1211109=11,88012 \cdot 11 \cdot 10 \cdot 9 = 11\text{,}880

Combinations

A combination is a selection of elements of a set such that order does not matter. The number of combinations of nn objects chosen rr at a time can be determined by a formula.

A combination is the selection of objects in groups where order does not matter. There are many ways that three different types of fruit can be selected from five types of fruit available at a farm stand. When order does not matter, selecting an apple, a peach, and strawberries versus selecting strawberries, a peach, and an apple results in the same three different types of fruits. The order in which the different types of fruit are selected does not matter, so the situation involves combinations.

The number of combinations of nn objects chosen rr at a time is:
C(n,r)=n!(nr)!r!C(n,r) = \frac {n!}{(n-r)!r!}
Note that each object in a combination can be chosen only once.
Step-By-Step Example
Using the Combination Formula
Out of a group of 80 students, 6 students will be selected to be on the dean's advisory council. How many different ways are there for selecting 6 students to be on the dean's advisory council?
Step 1

Determine whether the situation involves permutations or combinations. Then find the values for nn and rr.

Each student will have the same role on the council, so the order in which the students are selected does not matter. The situation involves combinations.

There are n=80n=80 students vying to be on the council. There are r=6r=6 students who will be selected to be on the council.

Step 2
Substitute values in the combination formula and simplify.
C(n,r)=n!(nr)!r!C(80,6)=80!(806)!6!=80797877767574!74!6!=8079787776756!=807978777675654321=300,500,200\begin{aligned}C(n,r) &= \frac {n!}{(n-r)!r!} \\ C(80,6) &= \frac {80!}{(80-6)!6!}\\ &= \frac {80 \cdot 79 \cdot 78 \cdot 77 \cdot 76 \cdot 75 \cdot {\color{#c42126}74!}}{{\color{#c42126}74!}6!} \\ &= \frac{80 \cdot 79 \cdot 78 \cdot 77 \cdot 76 \cdot 75}{6!}\\ &= \frac{80 \cdot 79 \cdot 78 \cdot 77 \cdot 76 \cdot 75}{6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}\\ &=300\text{,}500\text{,}200\end{aligned}
Solution
There are 300,500,200 different ways to choose 6 students to serve on the dean's advisory council.
Step-By-Step Example
Choosing Which Formula to Use
How many ways can five letters of the word "discovery" be rearranged?
Step 1

Determine whether the situation involves permutations or combinations. Then find the values for nn and rr.

Rearranging the letters means putting them in a different order, so the order of the letters matters. The situation involves permutations.

There are n=9n=9 letters in the word. There are r=5r=5 letters that will be rearranged for each permutation.

Step 2

Some of the letters are being selected and put into a different order.

Substitute values in the permutation formula and simplify.
P(n,r)=n!(nr)!P(9,5)=9!(95)!=987654!4!=98765=15,120\begin{aligned}P(n,r) &= \frac {n!}{(n-r)!} \\P(9,5) &= \frac {9!}{(9-5)!}\\&= \frac {9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot {\color{#c42126}4!}}{{\color{#c42126}4!}}\\&= 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5\\&= 15\text{,}120\end{aligned}
Solution
There are 15,120 different arrangements of five letters from the word "discovery."