Factorials and Ordered Selections

Why Order Changes the Count

When a set of distinct objects is placed into a sequence, changing their order usually creates a different outcome. For example, placing three separately labeled inspection instruments in the order A, B, C is not the same arrangement as placing them in the order B, A, C. Counting these sequences is a basic use of the multiplication principle: fill positions one at a time, counting the available choices at each step.

Factorial notation packages the resulting products into a compact form. It is useful both when every object is arranged and when only some objects are selected and placed into ordered positions. The central question is whether two outcomes that contain the same objects but list them in different orders count as distinct. In this tutorial, they do.

Definition: For a positive integer $n$, the factorial of $n$ is the product of the positive integers from $n$ down to 1: $$ n! = n(n-1)(n-2)\cdots 2\cdot 1. $$ By convention, $0!=1$. Factorial notation is used for nonnegative integers; it is not a count of arrangements of a negative number of objects.

For instance, $5!=5\cdot4\cdot3\cdot2\cdot1=120$, while $3!=3\cdot2\cdot1=6$. The convention $0!=1$ makes factorial formulas work at the boundary where no objects remain to be arranged. There is exactly one way to do nothing: leave the empty sequence as it is.

Arranging Every Object

Suppose $n$ distinct objects are to be arranged in $n$ ordered positions, with each object used exactly once. There are $n$ choices for the first position. Once it is filled, $n-1$ objects remain for the second position; then $n-2$ remain for the third, and so on, until one object fills the last position. By the multiplication principle, the total number of arrangements is $n(n-1)\cdots1=n!$.

Theorem: The number of arrangements of all $n$ distinct objects in a sequence is $n!$. This result assumes that every object is distinguishable, every object is used exactly once, and different orders count as different outcomes.

Worked Example: Ordering a Set of Distinct Inspection Labels

A hypothetical quality-control team has five separately labeled inspection cards, one for each station: A, B, C, D, and E. The cards are placed in a row to determine the order in which the stations are reviewed. How many possible review orders are there?

Method. Count the available choices for each successive position. Each card is distinct, each is used once, and a different sequence is a different review order.

There are 5 choices for the first position, 4 remaining choices for the second, 3 for the third, 2 for the fourth, and 1 for the last. Thus, $$ 5\cdot4\cdot3\cdot2\cdot1=5!=120. $$

There are 120 possible review orders. As a check, the factorial expands to exactly the same five position-by-position factors. A common mistake would be to count only the five cards and report 5; that counts the objects, not the different sequences they can form.

Ordering Only Part of a Collection

Now suppose there are $n$ distinct objects but only $r$ ordered positions to fill, where $0\le r\le n$. The first position has $n$ possible objects. After one is used, the next has $n-1$ choices, and the process continues for $r$ positions. No object can be reused.

Definition: An ordered selection of $r$ objects from $n$ distinct objects is a sequence of length $r$ whose entries are different objects from the collection. Its number is $$ n(n-1)(n-2)\cdots(n-r+1)=\frac{n!}{(n-r)!}. $$ The formula applies when $n$ and $r$ are integers with $0\le r\le n$, and when order matters and repetition is not allowed.

The notation $n!/(n-r)!$ is obtained by writing out $n!$ and cancelling the factors below $n-r+1$. For example, when $n=8$ and $r=3$, $$ \frac{8!}{(8-3)!}=\frac{8!}{5!} =\frac{8\cdot7\cdot6\cdot5!}{5!} =8\cdot7\cdot6. $$ The denominator is therefore the factorial of the number of objects left unused, not the factorial of the number of positions filled.

Why the Ordered-Selection Formula Works

The factorial ratio follows directly from the choices available at each position. The full factorial $n!$ contains all descending factors down to 1. But an ordered selection that fills only $r$ positions uses only the first $r$ factors, from $n$ down to $n-r+1$. The remaining factors multiply to $(n-r)!$, so dividing by them leaves precisely the factors needed.

Count choices by position.
The first of $r$ positions has $n$ choices. With no repetition, each later position has one fewer available object.

Apply the multiplication principle.
Multiplying the choices gives $n(n-1)\cdots(n-r+1)$. Each choice sequence determines exactly one ordered selection.

Rewrite using factorials.
Expand $n!=n(n-1)\cdots(n-r+1)(n-r)!$, then divide by $(n-r)!$. The result is $n!/(n-r)!$, as claimed.

This argument also checks the endpoint cases. If $r=n$, then $n!/(n-n)!=n!/0!=n!$, the count for arranging every object. If $r=0$, the product has no choice factors and represents the single empty sequence; the ratio is $n!/n!=1$.

Worked Example: Ranking Three Applicants

A hypothetical training program has eight distinct applicants. It will record first, second, and third place in a skills assessment. How many different ordered rankings of the top three are possible?

Method. This is an ordered selection without replacement: the position in the ranking matters, and an applicant cannot occupy more than one place.

There are 8 choices for first place, 7 remaining choices for second, and 6 remaining choices for third. Therefore, $$ 8\cdot7\cdot6=336. $$ Using factorial notation gives the same calculation: $$ \frac{8!}{(8-3)!}=\frac{8!}{5!} =\frac{8\cdot7\cdot6\cdot5!}{5!} =8\cdot7\cdot6=336. $$

There are 336 possible rankings. The result can be checked by noting that the first-place choice has 8 options and each of those can be followed by $7\cdot6=42$ possible second- and third-place sequences; $8\cdot42=336$. The common mistake to avoid is using $8!$, which counts complete rankings of all eight applicants rather than rankings of only the top three.

Reading and Simplifying Factorial Expressions

Factorial ratios are often easier to evaluate by cancelling before multiplying. Expanding both factorials in full can create large numbers that are unnecessary and can increase arithmetic errors. In $n!/(n-r)!$, expand only the numerator far enough to expose the denominator as a common factor.

Worked Example: Assigning Distinct Positions

A hypothetical review panel has nine distinct members. Four different roles—chair, recorder, presenter, and timekeeper—must each be assigned to a different member. How many role assignments are possible?

Method. Treat the four named roles as four ordered positions. The labels make the positions distinct, so swapping two members between roles changes the assignment. Each member can receive at most one role.

There are 9 choices for chair, then 8 for recorder, 7 for presenter, and 6 for timekeeper. The count is $$ 9\cdot8\cdot7\cdot6=3024. $$ The factorial expression confirms it: $$ \frac{9!}{(9-4)!}=\frac{9!}{5!} =\frac{9\cdot8\cdot7\cdot6\cdot5!}{5!} =9\cdot8\cdot7\cdot6=3024. $$

Thus, 3,024 role assignments are possible. A useful check is that $9!=362{,}880$ and $5!=120$, so $362{,}880/120=3{,}024$. The direct product and the factorial ratio agree. The mistake avoided here is to treat the roles as interchangeable: chair and recorder are different positions, so exchanging their assigned members produces another assignment.

Common Counting Traps

The most important decision is how the outcome is defined. If a result records a sequence, ranking, schedule, or assignment to named positions, order matters. If changing the order does not change the outcome, the ordered-selection formula is not the appropriate count; unordered selections are treated separately in the next subject.

Using $n!$ for a partial sequence. $n!$ counts arrangements using all $n$ objects. For only $r$ positions, use $n!/(n-r)!$.
Putting the wrong factorial in the denominator. The unused objects number $n-r$, so the denominator is $(n-r)!$, not $r!$.
Forgetting that order matters. If positions are labeled or ranked, swapping two selected objects changes the outcome.
Allowing repetition by accident. The formula assumes an object used in one position is no longer available for another. A model that allows reuse has different position counts.
Applying factorials outside their domain. In this setting $n$ and $r$ are nonnegative integers, with $r\le n$.

Common Mistake: Dividing by the Number of Positions

Suppose three of eight distinct objects are to be placed in a first, second, and third position. A mistaken calculation is $$ \frac{8!}{3!}=\frac{40{,}320}{6}=6{,}720. $$ This is not the number of ordered selections. The denominator $3!$ removes a factor that should remain: changing which object is first, second, or third changes the outcome.

The correct denominator counts the unused objects, of which there are $8-3=5$: $$ \frac{8!}{(8-3)!}=\frac{40{,}320}{120}=336. $$ This agrees with the position-by-position count $8\cdot7\cdot6=336$. The general correction is to identify how many objects remain unused and cancel that factorial.

Key takeaway: For $n$ distinct objects, arranging all of them gives $n!$ sequences. Filling $r$ ordered positions without repetition gives $n!/(n-r)!$. Before calculating, establish whether positions are distinct, whether every object is used, and whether reuse is allowed.

Check Your Understanding

For each situation, decide whether it is an arrangement of all objects or an ordered selection of part of the collection. Then calculate the count.

Six distinct route cards are placed in a complete sequence, using every card once. How many sequences are possible? (Answer: $6!=720$)
From ten distinct proposals, a panel records the first and second proposals to be reviewed, with no proposal repeated. How many ordered records are possible? (Answer: $10\cdot9=90$)
Six distinct volunteers are available for four different named duties, with one volunteer assigned to each duty. How many assignments are possible? (Answer: $6\cdot5\cdot4\cdot3=360$)
Evaluate $8!/5!$, and interpret the result as the number of ordered selections of objects from a collection of eight. (Answer: $8\cdot7\cdot6=336$, the number of ordered selections of three from eight)
A scheduler places three of nine distinct presentations into first, second, and third slots. Write the count as a factorial ratio and evaluate it. (Answer: $9!/6!=9\cdot8\cdot7=504$)

What You'll Learn