The Central Idea
Suppose \(5\) distinct people are available for three different offices: president, vice president, and secretary.
There are \(5\) choices for president. After that person is chosen, only \(4\) remain for vice president. Then \(3\) remain for secretary.
\[ 5\cdot4\cdot3=60. \]The offices are different, so changing who holds which office creates a new outcome.
This is a permutation without repetition.
What Is a Permutation?
A permutation is an ordered arrangement or ordered selection.
The word ordered is essential.
For example, the arrangements
\[ ABC \]and
\[ BAC \]use the same three objects, but they are different permutations because the order has changed.
What Does Without Repetition Mean?
Without repetition means that once an object has been selected, it cannot be selected again in the same arrangement.
If the available objects are
\[ A,\ B,\ C,\ D, \]then
\[ ABC \]is allowed, but
\[ ABA \]is not, because \(A\) has been used twice.
Order Matters
A quick test for a permutation problem is to ask:
Would changing the order create a different outcome?
If the answer is yes, then order matters.
For example, first place, second place, and third place are different positions. If Alice finishes first and Ben finishes second, that is a different result from Ben finishing first and Alice finishing second.
The Slot Model
Ordered selections can be visualized as labeled slots.
Suppose \(7\) people compete for three positions:
\[ \boxed{\text{1st}}\qquad \boxed{\text{2nd}}\qquad \boxed{\text{3rd}}. \]The first slot has \(7\) choices.
Because repetition is not allowed, the second slot has \(6\) choices.
The third slot has \(5\) choices.
\[ 7\cdot6\cdot5=210. \]The Permutation Formula
If \(n\) distinct objects are available and \(r\) of them are selected and arranged in order without repetition, the number of permutations is
\[ \boxed{P(n,r)=\frac{n!}{(n-r)!}}. \]Another common notation is
\[ {}_nP_r. \]Both notations describe the same count.
Why the Formula Works
The first selected position has \(n\) choices.
The second has \(n-1\).
The third has \(n-2\).
Continuing until \(r\) positions have been filled gives
\[ n(n-1)(n-2)\cdots(n-r+1). \]Now expand \(n!\) only far enough to expose \((n-r)!\):
\[ n! = n(n-1)(n-2)\cdots(n-r+1)(n-r)!. \]Dividing by \((n-r)!\) leaves exactly the required product:
\[ \boxed{P(n,r)=\frac{n!}{(n-r)!}}. \]Example: Three Positions from Eight People
Eight people are candidates for president, vice president, and treasurer. No person may hold more than one office.
Here,
\[ n=8,\qquad r=3. \]Therefore,
\[ P(8,3)=\frac{8!}{5!}. \]Expand only as far as \(5!\):
\[ \frac{8\cdot7\cdot6\cdot5!}{5!} = 8\cdot7\cdot6. \]Thus,
\[ \boxed{P(8,3)=336}. \]Sequential Choices and the Formula Agree
The same problem can be solved directly by the multiplication principle:
\[ 8\cdot7\cdot6=336. \]Or by the permutation formula:
\[ \frac{8!}{(8-3)!} = \frac{8!}{5!} = 336. \]The formula is not a separate counting idea. It is a compact expression of sequential choices without replacement.
Full Permutations
If all \(n\) available objects are arranged, then \(r=n\).
\[ P(n,n)=\frac{n!}{(n-n)!}. \]Since
\[ (n-n)!=0!=1, \]we obtain
\[ \boxed{P(n,n)=n!}. \]Thus ordinary factorial arrangements are a special case of permutations.
Example: Arranging Six Books
Six distinct books are arranged on a shelf.
Since all six books are used,
\[ P(6,6)=6!. \]Therefore,
\[ \boxed{6!=720} \]arrangements are possible.
Partial Permutations
A partial permutation uses only some of the available objects.
Suppose \(10\) runners compete, but only the first four finishing positions are recorded.
The number of ordered outcomes is
\[ P(10,4) = 10\cdot9\cdot8\cdot7. \]Therefore,
\[ \boxed{P(10,4)=5040}. \]Why We Stop after \(r\) Factors
In \(P(n,r)\), only \(r\) positions are being filled.
For example,
\[ P(9,4) \]contains four descending factors:
\[ \boxed{9\cdot8\cdot7\cdot6}. \]We do not continue to \(5\cdot4\cdot3\cdot2\cdot1\), because the remaining objects are not being assigned to positions.
Counting Ordered Codes without Repetition
Suppose a three-character code is formed from \(8\) distinct symbols, and no symbol may be repeated.
Because the first, second, and third positions are distinct, order matters.
The count is
\[ P(8,3) = 8\cdot7\cdot6 = \boxed{336}. \]Example: Awarding Gold, Silver, and Bronze
Twelve athletes compete for gold, silver, and bronze medals with no ties.
The medal positions are distinct, so order matters.
\[ P(12,3) = 12\cdot11\cdot10. \]Therefore,
\[ \boxed{P(12,3)=1320}. \]Example: Choosing a Four-Digit Sequence
How many four-digit sequences can be formed from the digits \(1,2,3,4,5,6,7\) if no digit may repeat?
There are \(7\) choices for the first position, then \(6\), then \(5\), then \(4\).
\[ 7\cdot6\cdot5\cdot4. \]Thus,
\[ \boxed{P(7,4)=840}. \]Permutation Problems with Leading-Zero Restrictions
Some digit problems require extra care because the first position may have a restriction.
Suppose a three-digit number is formed from the digits \(0,1,2,3,4,5\) without repetition.
A three-digit number cannot begin with \(0\), so the first digit has only \(5\) choices:
\[ 1,2,3,4,5. \]After the first digit is chosen, \(5\) digits remain for the second position and \(4\) for the third.
\[ 5\cdot5\cdot4=\boxed{100}. \]This problem is permutation-like, but the restriction means the plain formula \(P(6,3)\) cannot be applied without adjustment.
Restrictions Can Change the First Factor
The permutation formula assumes every available object is initially allowed in every position.
If a position has a special restriction, count that position according to the actual number of legal choices.
Then continue with the remaining legal choices.
Example: A Required First Position
Seven students are to give presentations in a row. Ava must present first. How many schedules are possible?
Ava's position is fixed, so only the remaining six students must be arranged.
\[ 6!. \]Therefore,
\[ \boxed{720} \]schedules are possible.
Example: One Person Cannot Be First
Six people are arranged in a line, but Dana cannot be first.
The first position has \(5\) legal choices.
After the first person is selected, the remaining five people can be arranged freely:
\[ 5!. \]Therefore,
\[ 5\cdot5! = 5\cdot120 = \boxed{600}. \]Alternative Method: Count All and Subtract
The previous problem can also be solved by complementary counting.
All arrangements:
\[ 6!=720. \]Arrangements with Dana first:
\[ 5!=120. \]Therefore,
\[ 720-120=\boxed{600}. \]Both methods agree.
Example: Two Fixed Positions
Eight people are arranged in a row. Alex must be first and Morgan must be last.
Those two positions are fixed.
The remaining six people can be arranged in the six middle positions:
\[ 6!. \]Therefore,
\[ \boxed{720} \]arrangements satisfy the restriction.
When Objects Must Stay Together
Suppose six people are arranged in a row, and two particular people must stand next to each other.
Treat the pair temporarily as one block.
Then there are five units to arrange:
\[ 5!. \]The two people inside the block can switch places in
\[ 2! \]ways.
Therefore,
\[ 5!\cdot2! = 120\cdot2 = \boxed{240}. \]The Block Method
When several specified objects must remain together:
- treat the required group as a single unit;
- arrange the resulting units;
- arrange the objects inside the block;
- multiply the counts.
This is still an application of sequential counting and permutations.
Example: Three People Together
Seven people are arranged in a row. Three particular people must remain together.
Treat the three-person group as one block. Together with the other four people, there are
\[ 5 \]units to arrange.
\[ 5! \]The three people inside the block can be ordered in
\[ 3! \]ways.
Therefore,
\[ 5!\cdot3! = 120\cdot6 = \boxed{720}. \]When Two Objects Must Not Be Together
Complementary counting is often efficient when two particular objects must not be adjacent.
Suppose six people are arranged in a row, and Alex and Blair may not stand together.
Count all arrangements:
\[ 6!=720. \]Count arrangements in which Alex and Blair are together:
\[ 5!\cdot2!=240. \]Subtract:
\[ 720-240=\boxed{480}. \]Permutation versus Combination
The decisive question is whether order matters.
Selecting Alice, Ben, and Carmen as three committee members creates the same committee regardless of the order in which their names are listed.
But assigning Alice as president, Ben as vice president, and Carmen as secretary is different from assigning Ben as president, Alice as vice president, and Carmen as secretary.
Committees are usually combinations. Distinct offices are permutations.
A Useful Order Test
Imagine two selected objects switching positions.
If the outcome changes, order matters.
If the outcome does not change, order does not matter.
This test often reveals whether a permutation formula is appropriate.
Example: Officers versus Committee Members
Five people are available.
If two different offices are assigned, the count is
\[ P(5,2)=5\cdot4=20. \]But if two people are merely chosen to form a committee, the pair \(\{A,B\}\) is the same committee as \(\{B,A\}\).
That second problem should not be counted as a permutation.
Permutation Trees
A decision tree can visualize a permutation count.
Suppose three objects \(A,B,C\) are arranged two at a time.
From \(A\), the second position can be \(B\) or \(C\):
\[ AB,\ AC. \]From \(B\):
\[ BA,\ BC. \]From \(C\):
\[ CA,\ CB. \]There are
\[ \boxed{6} \]ordered selections, agreeing with
\[ P(3,2)=3\cdot2=6. \]Permutations as Injective Assignments
A permutation without repetition can also be viewed as assigning objects to labeled positions with no object assigned twice.
Each position receives one distinct object.
In more advanced language, this is an injective assignment from the set of positions into the set of available objects.
The elementary counting formula remains
\[ \boxed{P(n,r)=\frac{n!}{(n-r)!}}. \]Boundary Case: Selecting Nothing
What is
\[ P(n,0)? \]The formula gives
\[ P(n,0) = \frac{n!}{n!} = \boxed{1}. \]There is exactly one ordered selection of zero objects: the empty selection.
Boundary Case: Selecting One Object
If one position is filled from \(n\) distinct objects, then
\[ P(n,1) = \frac{n!}{(n-1)!} = \boxed{n}. \]This agrees with the obvious fact that there are \(n\) choices for one position.
Boundary Case: Selecting Every Object
If every object is selected,
\[ P(n,n) = \frac{n!}{0!} = n!. \]Because \(0!=1\),
\[ \boxed{P(n,n)=n!}. \]Why \(r\le n\)
Without repetition, we cannot select more distinct objects than are available.
Therefore, in ordinary permutation problems,
\[ 0\le r\le n. \]Asking for \(P(5,7)\) would mean trying to fill seven distinct selected positions using only five available objects without reusing any object.
That cannot be done.
Factorial Simplification in Permutations
Suppose we need
\[ P(15,3). \]Do not calculate \(15!\) and \(12!\) separately.
Instead,
\[ P(15,3) = \frac{15!}{12!} = 15\cdot14\cdot13. \]Therefore,
\[ \boxed{P(15,3)=2730}. \]Writing the Product Directly
For small \(r\), the direct descending product is often faster than writing the full factorial formula.
For example,
\[ P(20,4) \]can be written immediately as
\[ \boxed{20\cdot19\cdot18\cdot17}. \]The number of factors must equal \(r\).
Example: License-Style Sequences
Suppose a system creates sequences of four distinct letters chosen from \(26\) letters, with no repetition.
The number of possible sequences is
\[ P(26,4) = 26\cdot25\cdot24\cdot23. \]Therefore,
\[ \boxed{P(26,4)=358800}. \]Example: Ranking Finalists
A competition has \(9\) finalists. The judges rank the top \(5\) in order.
The number of possible rankings is
\[ P(9,5) = 9\cdot8\cdot7\cdot6\cdot5. \]Thus,
\[ \boxed{P(9,5)=15120}. \]Example: Assigning Distinct Jobs
Ten workers are available for four distinct one-person jobs.
Because the jobs are different, assigning workers to them is ordered.
\[ P(10,4) = 10\cdot9\cdot8\cdot7. \]Therefore,
\[ \boxed{5040} \]assignments are possible.
Example: Order Matters Even without Physical Arrangement
A permutation does not require objects to be physically placed in a row.
Suppose \(8\) candidates are available for three distinct leadership roles.
The roles themselves create the order:
\[ \text{Role 1},\quad \text{Role 2},\quad \text{Role 3}. \]Thus,
\[ P(8,3)=\boxed{336}. \]Common Error: Ignoring Order
Suppose first and second place are awarded from \(5\) competitors.
A student might count only the unordered pairs of winners.
But
\[ AB \]and
\[ BA \]represent different first-and-second-place outcomes.
The correct count is
\[ P(5,2)=5\cdot4=\boxed{20}. \]Common Error: Allowing Repetition
If repetition is not allowed, the number of choices decreases after each selection.
For three ordered positions chosen from \(6\) distinct objects, the count is
\[ 6\cdot5\cdot4, \]not
\[ 6\cdot6\cdot6. \]Therefore,
\[ \boxed{P(6,3)=120}. \]Common Error: Using \(n!\) When Only \(r\) Positions Matter
If \(10\) people compete for only three ranked positions, we do not arrange all ten people.
The count is
\[ P(10,3)=10\cdot9\cdot8, \]not
\[ 10!. \]Only three positions are being assigned.
Common Error: Using the Wrong Denominator
The formula is
\[ \boxed{P(n,r)=\frac{n!}{(n-r)!}}. \]The denominator is not \(r!\).
For example,
\[ P(7,3) = \frac{7!}{4!} = 7\cdot6\cdot5 = \boxed{210}. \]Common Error: Forgetting a Position Restriction
A formula should not override the conditions of the problem.
If the first position cannot contain a particular object, then the first factor must reflect that restriction.
Always count legal choices, not merely available objects.
Common Error: Confusing Objects with Positions
In \(P(n,r)\),
- \(n\) is the number of available distinct objects;
- \(r\) is the number of ordered positions being filled.
Keeping these roles separate prevents many formula errors.
A Reliable Procedure for Permutations without Repetition
- Identify the available distinct objects.
- Identify the positions or selected slots.
- Confirm that order matters.
- Confirm that an object cannot be reused.
- Set \(n\) equal to the number of available objects.
- Set \(r\) equal to the number of ordered selections.
- Use \(P(n,r)=\frac{n!}{(n-r)!}\) when all positions are unrestricted.
- Adjust individual factors when restrictions apply.
- Simplify factorials by cancellation before calculating.
- Check that the product contains exactly \(r\) descending factors.
Example: Full Workflow
Eleven students are available for four distinct offices.
Order matters because the offices are different.
Repetition is not allowed because one student cannot occupy more than one of the four offices.
Therefore,
\[ n=11,\qquad r=4. \]Apply the formula:
\[ P(11,4) = \frac{11!}{7!}. \]Expand only as far as \(7!\):
\[ \frac{11\cdot10\cdot9\cdot8\cdot7!}{7!}. \]Thus,
\[ \boxed{P(11,4)=11\cdot10\cdot9\cdot8=7920}. \]Practice 1: Basic Permutation
Evaluate:
\[ P(7,3). \]Practice 2: Ranked Finishers
Ten runners compete with no ties. How many possible outcomes are there for first, second, and third place?
Practice 3: Distinct Offices
Eight people are available for president, vice president, secretary, and treasurer. How many assignments are possible if no person may hold more than one office?
Practice 4: Full Arrangement
How many ways can \(7\) distinct objects be arranged in a row?
Practice 5: Ordered Code
How many four-character codes can be formed from \(9\) distinct symbols if no symbol may repeat?
Practice 6: Fixed First Position
Six speakers are scheduled in a row. Jordan must speak first. How many schedules are possible?
Practice 7: Two People Together
Five people are arranged in a row. Two particular people must stand together. How many arrangements are possible?
Practice 8: Two People Not Together
Five people are arranged in a row. Two particular people must not stand together. How many arrangements are possible?
Practice 9: Leading-Zero Restriction
How many three-digit numbers can be formed from \(0,1,2,3,4,5,6\) without repetition?
Practice 10: Identify the Error
A student says that the number of ways to award first and second place among six competitors is
\[ 6^2=36. \]Explain the error and give the correct count.
Practice Answers
Practice 1
\[ P(7,3) = 7\cdot6\cdot5 = \boxed{210}. \]Practice 2
\[ P(10,3) = 10\cdot9\cdot8 = \boxed{720}. \]Practice 3
\[ P(8,4) = 8\cdot7\cdot6\cdot5 = \boxed{1680}. \]Practice 4
\[ P(7,7)=7!=\boxed{5040}. \]Practice 5
\[ P(9,4) = 9\cdot8\cdot7\cdot6 = \boxed{3024}. \]Practice 6
Jordan is fixed first, so the remaining five speakers can be arranged in
\[ \boxed{5!=120} \]ways.
Practice 7
Treat the required pair as one block. There are four units to arrange:
\[ 4!. \]The two people inside the block can switch places:
\[ 2!. \]Therefore,
\[ 4!\cdot2! = 24\cdot2 = \boxed{48}. \]Practice 8
All arrangements:
\[ 5!=120. \]Arrangements with the two people together:
\[ 4!\cdot2!=48. \]Therefore,
\[ 120-48=\boxed{72}. \]Practice 9
The first digit has \(6\) choices because it cannot be \(0\).
Then \(6\) digits remain for the second position and \(5\) for the third.
\[ 6\cdot6\cdot5=\boxed{180}. \]Practice 10
The count \(6^2\) incorrectly allows the same competitor to occupy both places.
Without repetition, there are \(6\) choices for first place and \(5\) remaining choices for second:
\[ \boxed{6\cdot5=30}. \]What to Remember
A permutation is an ordered arrangement or ordered selection.
Without repetition, each selected object can be used at most once.
For \(n\) distinct available objects and \(r\) ordered selections,
\[ \boxed{P(n,r)=\frac{n!}{(n-r)!}}. \]Equivalently,
\[ \boxed{P(n,r)=n(n-1)(n-2)\cdots(n-r+1)}. \]The most important diagnostic question is: does changing the order create a different outcome?
If yes, and repetition is not allowed, a permutation model is often appropriate.
Where We Go Next
Permutations without repetition assume that each available object can be used at most once. Other ordered arrangements allow repeated symbols, repeated choices, or repeated object types.
The next lesson, Permutations with Repetition, examines how counting changes when repetition is allowed or when some objects are indistinguishable.