The Central Idea
Suppose \(6\) distinct people are arranged in a row.
Without restrictions, there are
\[ 6! \]arrangements.
Now suppose Alex must stand first.
Alex's position is fixed, so only the remaining five people must be arranged:
\[ \boxed{5!}. \]The restriction changes the set of legal choices. Good counting begins by identifying exactly what the restriction changes.
Start with the Unrestricted Model
Before handling a restriction, identify the unrestricted count.
For \(n\) distinct objects arranged in a row,
\[ n! \]arrangements are possible.
Restrictions either:
- fix some of those positions;
- remove some legal choices;
- force objects into blocks;
- force objects into gaps;
- exclude forbidden cases;
- divide the problem into cases.
Restriction Problems Are Structural
The same phrase can lead to different methods depending on the arrangement.
“Together” often suggests a block.
“Not together” often suggests a complement.
“No two adjacent” often suggests a gap method.
“Exactly one of these positions” often suggests cases.
“None in their original positions” often suggests inclusion–exclusion.
Fixed Positions
If a particular object must occupy a particular position, place it there first.
Then arrange the remaining objects in the remaining positions.
For \(n\) distinct objects with one specified object fixed in one specified position,
\[ \boxed{(n-1)!}. \]Example: One Fixed Position
Seven students give presentations. Dana must present first.
Fix Dana in the first position.
Arrange the remaining six students:
\[ \boxed{6!=720}. \]Several Fixed Positions
If \(k\) specified objects are fixed in \(k\) specified positions, only \(n-k\) objects remain to be arranged.
Therefore,
\[ \boxed{(n-k)!}. \]Example: First and Last Fixed
Eight people stand in a line. Alex must be first and Blair must be last.
Those two positions are fixed.
The remaining six people can be arranged in
\[ \boxed{6!=720} \]ways.
One of Several Allowed Positions
Sometimes an object is restricted to a small set of positions rather than one exact position.
If the cases are disjoint, count each allowed position separately and add.
Example: Alex Must Be at an End
Six people stand in a row. Alex must be either first or last.
Alex has
\[ 2 \]allowed positions.
For each choice, arrange the remaining five people:
\[ 5!. \]Therefore,
\[ 2\cdot5! = \boxed{240}. \]At Least One of Several Special Positions
If a restriction says an object must occupy one of several positions, multiplication is often appropriate when choosing the position and then arranging everything else.
But if several different objects may satisfy the condition in overlapping ways, case analysis or inclusion–exclusion may be required.
Forbidden Positions
If one particular object cannot occupy one particular position, a complement is often easier than direct counting.
Count all arrangements, then subtract arrangements with the object in the forbidden position.
Example: Dana Cannot Be First
Six people stand in a row. Dana cannot be first.
All arrangements:
\[ 6!=720. \]Arrangements with Dana first:
\[ 5!=120. \]Therefore,
\[ 720-120 = \boxed{600}. \]Direct Method for One Forbidden Position
The previous problem can also be counted directly.
The first position has
\[ 5 \]legal choices.
The remaining five people can be arranged in
\[ 5! \]ways.
Thus,
\[ 5\cdot5! = \boxed{600}. \]Choose the Easier View
Many restricted arrangements can be counted in more than one way.
The direct method counts legal arrangements.
The complementary method counts all arrangements and removes illegal ones.
Prefer the method that creates fewer cases and simpler arithmetic.
Two Objects Must Be Together
If two specified objects must be adjacent in a linear arrangement, treat them as a temporary block.
With \(n\) total objects, the pair plus the other \(n-2\) objects creates
\[ n-1 \]units.
Arrange those units in
\[ (n-1)! \]ways.
The two objects inside the block can switch in
\[ 2! \]ways.
Therefore,
\[ \boxed{2(n-1)!}. \]Example: Two People Together
Seven people stand in a row. Alex and Blair must stand together.
Treat Alex and Blair as one unit.
There are six units to arrange:
\[ 6!. \]Alex and Blair can appear as
\[ AB \]or
\[ BA. \]Therefore,
\[ 6!\cdot2 = \boxed{1440}. \]Three Objects Must Be Together
If three specified objects must remain consecutive, treat all three as one block.
With \(n\) total objects, the block plus the remaining \(n-3\) objects gives
\[ n-2 \]units.
Arrange the units in
\[ (n-2)! \]ways.
Arrange the three objects inside the block in
\[ 3! \]ways.
Therefore,
\[ \boxed{3!(n-2)!}. \]Example: Three Books Together
Eight distinct books are arranged on a shelf. Three particular books must remain together.
Treat the three books as one block.
The block plus the other five books gives six units:
\[ 6!. \]The three books inside the block can be ordered in
\[ 3!. \]Therefore,
\[ 6!\cdot3! = 720\cdot6 = \boxed{4320}. \]General Block Formula
If \(k\) specified distinct objects among \(n\) must remain consecutive, treat them as one block.
The number of external units is
\[ n-k+1. \]Their arrangements total
\[ (n-k+1)!. \]The \(k\) internal objects can be arranged in
\[ k!. \]Therefore,
\[ \boxed{k!(n-k+1)!}. \]Two Separate Blocks
If two disjoint groups must each remain together, treat each group as its own block.
Then count:
- the arrangements of the external units;
- the internal arrangements within each block.
Example: Two Pairs Must Stay Together
Eight people stand in a row. Alex and Blair must stay together, and Casey and Dana must stay together.
Treat the two pairs as two blocks.
Together with the remaining four people, there are
\[ 6 \]units.
Arrange them:
\[ 6!. \]Each pair can switch internally:
\[ 2!\cdot2!. \]Therefore,
\[ 6!\cdot2\cdot2 = \boxed{2880}. \]Two Objects Must Not Be Together
The complement method is usually simplest.
Count all arrangements, then subtract arrangements in which the specified objects are adjacent.
For \(n\) distinct objects,
\[ \boxed{n!-2(n-1)!}. \]Example: Two People Not Together
Seven people stand in a row. Alex and Blair must not stand together.
All arrangements:
\[ 7!=5040. \]Arrangements with Alex and Blair together:
\[ 6!\cdot2=1440. \]Therefore,
\[ 5040-1440 = \boxed{3600}. \]Objects Must Be Separated
If several objects must not be adjacent to each other, arranging the other objects first often creates useful gaps.
This is called the gap method.
The Gap Method
Suppose \(m\) ordinary objects are arranged in a row.
They create
\[ m+1 \]gaps:
\[ \_\,O\,\_\,O\,\_\,O\,\_\cdots\_. \]To keep special objects separated, place at most one special object in each gap.
Example: Three Special Objects Cannot Be Adjacent
Eight distinct people include three members of group \(A\) and five members of group \(B\). No two members of group \(A\) may stand together.
First arrange the five members of group \(B\):
\[ 5!. \]They create six gaps.
Choose three of those gaps:
\[ \binom{6}{3}. \]Arrange the three members of group \(A\) in the chosen gaps:
\[ 3!. \]Therefore,
\[ 5!\binom{6}{3}3! = 120\cdot20\cdot6 = \boxed{14400}. \]Why the Gap Method Prevents Adjacency
Once the ordinary objects are fixed, every chosen gap can hold at most one restricted object.
Because two restricted objects never occupy the same gap, they cannot become adjacent.
Feasibility of Separation
If \(s\) special objects must be separated by \(m\) ordinary objects, there are only
\[ m+1 \]available gaps.
Therefore, a necessary condition is
\[ \boxed{s\le m+1}. \]If this fails, the arrangement is impossible.
Example: Impossible Separation
Can six \(A\)-type objects be arranged with four \(B\)-type objects so that no two \(A\)'s are adjacent?
Four \(B\)'s create only
\[ 5 \]gaps.
But six \(A\)'s would require six distinct gaps.
Therefore, the arrangement is impossible.
Alternating Arrangements
If two groups must alternate, their sizes must differ by at most one in a linear arrangement.
If the group sizes are equal, there are usually two possible type patterns:
\[ ABAB\cdots \]or
\[ BABA\cdots. \]Example: Equal Groups Alternate
Four distinct members of group \(A\) and four distinct members of group \(B\) stand in a row, alternating by group.
There are two type patterns:
\[ ABABABAB \]and
\[ BABABABA. \]Arrange the \(A\)'s in their four positions:
\[ 4!. \]Arrange the \(B\)'s:
\[ 4!. \]Therefore,
\[ 2(4!)(4!) = \boxed{1152}. \]Groups Differ by One
If one group has exactly one more member than the other, the larger group must occupy both ends.
Therefore, the type pattern is forced.
Example: Alternation with One Extra
Five distinct members of group \(A\) and four distinct members of group \(B\) must alternate in a row.
The only possible type pattern is
\[ ABABABABA. \]Arrange the five \(A\)'s:
\[ 5!. \]Arrange the four \(B\)'s:
\[ 4!. \]Therefore,
\[ 5!\cdot4! = \boxed{2880}. \]At Least One Object between Two Others
A condition such as “Alex and Blair must have at least one person between them” is equivalent to saying they may not be adjacent.
Therefore, a complement of the adjacency count usually works.
Exactly One Object between Two Specified Objects
This condition is more specific.
Suppose \(n\) distinct objects are arranged in a row and \(A\) and \(B\) must have exactly one object between them.
Their positions differ by \(2\).
Count possible position pairs, assign \(A\) and \(B\) to those positions, then arrange the remaining objects.
Example: Exactly One Person Between Alex and Blair
Six people stand in a row. Exactly one person must stand between Alex and Blair.
The possible unordered position pairs are
\[ (1,3),\ (2,4),\ (3,5),\ (4,6). \]There are
\[ 4 \]such pairs.
Alex and Blair can switch within the pair:
\[ 2. \]Arrange the remaining four people:
\[ 4!. \]Therefore,
\[ 4\cdot2\cdot4! = \boxed{192}. \]Exactly \(k\) Objects between Two Specified Objects
In a row of \(n\) positions, if exactly \(k\) objects must lie between \(A\) and \(B\), their positions differ by
\[ k+1. \]The number of unordered position pairs is
\[ n-(k+1). \]Assign \(A\) and \(B\) in two orders and arrange the remaining \(n-2\) objects.
Therefore,
\[ \boxed{ 2[n-(k+1)](n-2)! }. \]Example: Exactly Two Between
Seven people stand in a row. Exactly two people must stand between Alex and Blair.
Their positions differ by \(3\).
The possible position pairs are
\[ (1,4),\ (2,5),\ (3,6),\ (4,7). \]There are \(4\) pairs.
Therefore,
\[ 4\cdot2\cdot5! = 8\cdot120 = \boxed{960}. \]Relative Order Restrictions
Some restrictions concern relative order rather than exact positions.
For example:
- Alex must appear before Blair;
- three specified people must appear in alphabetical order;
- one task must precede another.
Symmetry among internal orderings often makes these problems simple.
Example: Alex Before Blair
Seven distinct people stand in a row. Alex must appear before Blair.
Among all
\[ 7! \]arrangements, exactly half have Alex before Blair and half have Blair before Alex.
Therefore,
\[ \boxed{\frac{7!}{2}=2520}. \]Three Objects in a Fixed Relative Order
Suppose three specified objects \(A,B,C\) must appear in the relative order
\[ A\text{ before }B\text{ before }C, \]though other objects may appear between them.
Among the
\[ 3! \]possible relative orders of \(A,B,C\), exactly one is allowed.
Therefore, for \(n\) total distinct objects,
\[ \boxed{\frac{n!}{3!}}. \]Example: Three Tasks in Order
Eight distinct tasks are scheduled. Tasks \(A,B,C\) must occur in that relative order, but need not be consecutive.
Therefore,
\[ \frac{8!}{3!} = \frac{40320}{6} = \boxed{6720}. \]Relative Order versus Consecutive Order
These are different restrictions.
“\(A,B,C\) appear in that order” allows other objects between them.
“\(A,B,C\) appear consecutively in that order” treats
\[ ABC \]as one fixed block.
Always distinguish relative order from adjacency.
Consecutive in a Fixed Internal Order
If \(k\) specified objects must appear consecutively in one prescribed internal order, treat the entire sequence as one block.
There is no internal \(k!\) factor because the order is fixed.
Therefore, with \(n\) total objects,
\[ \boxed{(n-k+1)!}. \]Example: ABC Must Appear Consecutively
Seven distinct letters are arranged, and \(A,B,C\) must appear consecutively in exactly the order
\[ ABC. \]Treat \(ABC\) as one block.
Together with the other four letters, there are five units.
Therefore,
\[ \boxed{5!=120}. \]At Least One of Several Objects at an End
Restrictions involving “at least one” often invite complementary counting.
Count all arrangements and subtract those in which none of the specified objects occupies an end.
Example: At Least One of Alex or Blair at an End
Six people stand in a row. At least one of Alex or Blair must occupy an end.
All arrangements:
\[ 6!=720. \]Count arrangements with neither Alex nor Blair at an end.
Choose and order two of the other four people at the two ends:
\[ P(4,2)=12. \]Arrange the remaining four people in the middle:
\[ 4!=24. \]Forbidden arrangements:
\[ 12\cdot24=288. \]Therefore,
\[ 720-288 = \boxed{432}. \]Exactly One of Two Objects at an End
Suppose exactly one of \(A\) and \(B\) must occupy an end.
Choose which one occupies an end, choose which end, then ensure the other special object is not placed at the opposite end.
This naturally becomes a sequence of restricted choices.
Example: Exactly One of Alex or Blair at an End
Six people stand in a row. Exactly one of Alex or Blair must occupy an end.
Choose which special person is at an end:
\[ 2. \]Choose which end:
\[ 2. \]The opposite end must be occupied by one of the other four people:
\[ 4. \]Arrange the remaining three middle positions:
\[ 3!. \]Therefore,
\[ 2\cdot2\cdot4\cdot3! = \boxed{96}. \]Several Restricted Positions
If multiple objects each have forbidden positions, direct sequential counting may become difficult because the number of legal choices depends on earlier choices.
In those situations, inclusion–exclusion is often more reliable.
A First Inclusion–Exclusion Arrangement
Suppose four people \(A,B,C,D\) occupy four labeled seats, and neither \(A\) nor \(B\) may sit in their own designated seats.
Start with all arrangements:
\[ 4!. \]Let \(E_A\) be the arrangements with \(A\) in the forbidden seat, and \(E_B\) the arrangements with \(B\) in the forbidden seat.
Then
\[ |E_A|=3!, \qquad |E_B|=3!. \]Their intersection fixes both:
\[ |E_A\cap E_B|=2!. \]Therefore, legal arrangements total
\[ 4!-3!-3!+2! = 24-6-6+2 = \boxed{14}. \]Derangements
A derangement is a permutation in which no object remains in its original designated position.
The number of derangements of \(n\) distinct objects is often written
\[ !n \]or
\[ D_n. \]Inclusion–exclusion gives
\[ \boxed{ D_n = n! \sum_{k=0}^{n}\frac{(-1)^k}{k!} }. \]This is an important restricted-arrangement model, though a full treatment belongs with broader inclusion–exclusion methods.
Example: Four-Letter Derangement
Four letters are placed into four addressed envelopes, and no letter may go into its matching envelope.
The derangement count is
\[ D_4 = 4! \left( 1-1+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!} \right). \]Therefore,
\[ \boxed{D_4=9}. \]Exactly \(k\) Objects Fixed
Suppose exactly \(k\) of \(n\) objects must remain in their original positions.
First choose which \(k\) objects remain fixed:
\[ \binom{n}{k}. \]The remaining \(n-k\) objects must be deranged:
\[ D_{n-k}. \]Therefore,
\[ \boxed{ \binom{n}{k}D_{n-k} }. \]Example: Exactly Two Fixed Positions
Five objects are permuted. How many permutations leave exactly two objects in their original positions?
Choose the two fixed objects:
\[ \binom{5}{2}=10. \]Derange the remaining three:
\[ D_3=2. \]Therefore,
\[ 10\cdot2 = \boxed{20}. \]Restrictions with Repeated Objects
If some arranged objects are identical, both the restriction and the repeated-object symmetry must be handled.
Often the simplest method is to impose the positional restriction first and then count the remaining repeated-object arrangements.
Example: Repeated Letters with a Fixed First Letter
Arrange the letters of
\[ \text{LEVEL} \]with \(L\) first.
Fix one \(L\).
The remaining letters are
\[ L,E,E,V. \]Therefore,
\[ \frac{4!}{2!} = \boxed{12}. \]Example: Repeated Letters Not Beginning with a Particular Type
Arrange the letters
\[ A,A,B,B,C \]so that the arrangement does not begin with \(C\).
All distinguishable arrangements:
\[ \frac{5!}{2!2!}=30. \]Arrangements beginning with \(C\):
\[ \frac{4!}{2!2!}=6. \]Therefore,
\[ 30-6 = \boxed{24}. \]Restrictions on Circular Arrangements
The same methods also appear in circular arrangements, but the unrestricted circular count must be used first.
For \(n\) distinct objects around an unlabeled circle,
\[ (n-1)! \]is the base count.
Blocks and complements must then be adjusted to circular rather than linear structure.
Example: Circular Adjacency
Six people sit around a round table. Alex and Blair must sit together.
Treat them as one block.
This creates five circular units, which can be arranged in
\[ 4! \]ways.
The pair can switch internally:
\[ 2. \]Therefore,
\[ 4!\cdot2 = \boxed{48}. \]Linear and Circular Blocks Are Different
In a row, a two-person block among \(n\) people creates \(n-1\) linear units and contributes
\[ (n-1)!. \]Around a circle, the same block creates \(n-1\) circular units and contributes
\[ (n-2)!. \]Always preserve the geometry of the original arrangement.
Counting by Cases
When a restriction naturally separates into disjoint possibilities, count each case and add.
A good case split should satisfy two conditions:
- every legal arrangement appears in some case;
- no legal arrangement appears in more than one case.
Example: A Must Be First or B Must Be Last
Six people stand in a row. Count arrangements in which Alex is first or Blair is last.
Arrangements with Alex first:
\[ 5!. \]Arrangements with Blair last:
\[ 5!. \]Arrangements satisfying both conditions have been counted twice:
\[ 4!. \]Therefore,
\[ 5!+5!-4! = 120+120-24 = \boxed{216}. \]Why Overlap Matters
If cases overlap, simply adding their counts overcounts outcomes lying in the intersection.
This is the simplest form of inclusion–exclusion:
\[ \boxed{ |A\cup B| = |A|+|B|-|A\cap B| }. \]Exactly One of Two Conditions
If exactly one of conditions \(A\) and \(B\) must hold, then count
\[ (A\text{ only})+(B\text{ only}). \]Equivalently,
\[ |A|+|B|-2|A\cap B|. \]The intersection must be removed from both condition counts.
Restrictions That Affect the First Choice
Sometimes a restriction can be handled by changing the first factor in a sequential product.
For example, a four-digit number formed without repetition from
\[ 0,1,2,3,4,5 \]cannot begin with \(0\).
The first digit has \(5\) choices, then \(5\), then \(4\), then \(3\).
Therefore,
\[ 5\cdot5\cdot4\cdot3 = \boxed{300}. \]Restrictions Can Change Later Choices Too
A positional restriction may affect more than one stage of a sequential count.
Always recompute the number of legal choices after earlier choices have been made.
Do not apply a memorized permutation formula when its assumptions no longer match the problem.
Arrange the More Restricted Group First
In many problems, one group has stronger placement restrictions than the other.
A useful heuristic is:
arrange the less restricted objects first when doing so creates positions or gaps for the more restricted objects.
This is the logic behind many separation and alternation problems.
Blocks and Gaps Are Dual Tools
A block method pulls objects together.
A gap method keeps objects apart.
These are complementary structural tools:
- together \(\rightarrow\) block;
- separated \(\rightarrow\) gaps.
Complementary Counting as a Restriction Tool
A condition such as “not adjacent,” “at least one,” or “not all” often has an easier complement.
The general pattern is
\[ \boxed{ \text{legal} = \text{all} - \text{illegal} }. \]Example: At Least One Vowel at an End
Suppose six distinct letters contain two vowels and four consonants. Count arrangements with at least one vowel at an end.
All arrangements:
\[ 6!. \]Count arrangements with consonants at both ends.
Choose and order two consonants for the ends:
\[ P(4,2)=12. \]Arrange the remaining four letters in the middle:
\[ 4!=24. \]Therefore,
\[ 6!-12\cdot24 = 720-288 = \boxed{432}. \]Common Error: Forgetting Internal Block Order
If Alex and Blair must stay together, the block can be
\[ AB \]or
\[ BA. \]Treating the pair as one unit counts only the external block placement.
Multiply by
\[ 2! \]unless the internal order is fixed.
Common Error: Multiplying Overlapping Cases
Multiplication applies when choices occur together as stages of one outcome.
If the problem instead gives alternative cases, those counts are generally added.
Always distinguish “and” from disjoint “or.”
Common Error: Adding Overlapping Cases without Correction
If condition \(A\) and condition \(B\) can both occur in the same arrangement, then
\[ |A|+|B| \]counts the intersection twice.
Use
\[ |A|+|B|-|A\cap B|. \]Common Error: Using a Block for “Not Together”
A block directly counts adjacency.
If two objects must not be together, the block usually counts the forbidden arrangements.
Subtract that count from the unrestricted total.
Common Error: Forgetting Edge Gaps
\(m\) arranged objects create
\[ m+1 \]gaps, not \(m-1\).
The two outside gaps count:
\[ \_\,O\,\_\,O\,\_\cdots O\,\_. \]Common Error: Ignoring Feasibility
Before counting a restricted arrangement, check whether the restriction can be satisfied at all.
For example, too many special objects cannot be separated if too few gaps are available.
An impossible arrangement has count
\[ \boxed{0}. \]Common Error: Confusing Relative Order with Fixed Positions
If Alex must appear before Blair, neither person is assigned a specific position.
The condition restricts only their relative order.
Fixing exact positions would be unnecessarily restrictive and produce an undercount.
Common Error: Treating Relative Order as Consecutive
“\(A\) before \(B\)” allows any number of objects between them.
“\(A\) immediately before \(B\)” requires the block
\[ AB. \]These restrictions have different counts.
Common Error: Using Linear Formulas for Circular Arrangements
A round-table problem has rotational equivalence.
The unrestricted base count is
\[ (n-1)!, \]not
\[ n!. \]Restriction methods must preserve that circular structure.
A Reliable Restriction Checklist
- Write the unrestricted count.
- Identify whether the arrangement is linear or circular.
- Mark fixed positions.
- Mark forbidden positions.
- Identify required adjacency.
- Identify required separation.
- Identify relative-order conditions.
- Check for repeated objects.
- Check for overlapping cases.
- Check feasibility before calculating.
A Reliable Method Map
- fixed position \(\rightarrow\) place first;
- allowed position set \(\rightarrow\) choose position, then arrange;
- must be together \(\rightarrow\) block method;
- must not be together \(\rightarrow\) complement of block count;
- no two from a group adjacent \(\rightarrow\) gap method;
- groups alternate \(\rightarrow\) type pattern plus internal permutations;
- relative order fixed \(\rightarrow\) divide by internal order count;
- overlapping conditions \(\rightarrow\) inclusion–exclusion;
- none in original position \(\rightarrow\) derangement methods.
Example: Full Block Workflow
Nine people stand in a row. Three particular people must remain together.
Treat the three as one block.
The block plus the other six people creates
\[ 7 \]external units.
Arrange the units:
\[ 7!. \]Arrange the three people inside the block:
\[ 3!. \]Therefore,
\[ 7!\cdot3! = 5040\cdot6 = \boxed{30240}. \]Example: Full Gap Workflow
Nine people consist of four members of group \(A\) and five members of group \(B\). No two members of group \(A\) may stand together.
Arrange the five \(B\)'s:
\[ 5!. \]They create six gaps.
Choose four gaps:
\[ \binom{6}{4}. \]Arrange the four \(A\)'s in those gaps:
\[ 4!. \]Therefore,
\[ 5!\binom{6}{4}4! = 120\cdot15\cdot24 = \boxed{43200}. \]Example: Full Complement Workflow
Eight people stand in a row. Alex and Blair must not be adjacent.
Count all arrangements:
\[ 8!. \]Count forbidden adjacent arrangements:
\[ 7!\cdot2. \]Subtract:
\[ 8!-2(7!) = 40320-10080 = \boxed{30240}. \]Practice 1: Fixed Position
Seven people stand in a row. Alex must be first. How many arrangements are possible?
Practice 2: Forbidden Position
Seven people stand in a row. Alex cannot be first. How many arrangements are possible?
Practice 3: Pair Together
Eight people stand in a row. Alex and Blair must stand together. How many arrangements are possible?
Practice 4: Pair Apart
Eight people stand in a row. Alex and Blair must not stand together. How many arrangements are possible?
Practice 5: Three Together
Nine people stand in a row. Three particular people must remain consecutive. How many arrangements are possible?
Practice 6: No Two Special Objects Adjacent
Seven distinct people include three members of group \(A\) and four members of group \(B\). No two members of group \(A\) may stand together. How many arrangements are possible?
Practice 7: Alternating Groups
Three distinct members of group \(A\) and three distinct members of group \(B\) must alternate in a row. How many arrangements are possible?
Practice 8: Relative Order
Seven distinct people stand in a row. Alex must appear before Blair. How many arrangements are possible?
Practice 9: Exactly One Person Between
Seven people stand in a row. Exactly one person must stand between Alex and Blair. How many arrangements are possible?
Practice 10: Identify the Error
A student says that if two specified people among seven must stand together, the number of arrangements is
\[ 6!. \]Explain what has been omitted and give the correct count.
Practice Answers
Practice 1
Fix Alex and arrange the remaining six:
\[ \boxed{6!=720}. \]Practice 2
Count all arrangements and subtract those with Alex first:
\[ 7!-6! = 5040-720 = \boxed{4320}. \]Practice 3
Treat Alex and Blair as one block:
\[ 7!\cdot2 = 5040\cdot2 = \boxed{10080}. \]Practice 4
Subtract adjacent arrangements from all arrangements:
\[ 8!-2(7!) = 40320-10080 = \boxed{30240}. \]Practice 5
Treat the three people as one block:
\[ 7!\cdot3! = 5040\cdot6 = \boxed{30240}. \]Practice 6
Arrange the four \(B\)'s:
\[ 4!. \]They create five gaps.
Choose three gaps and arrange the three \(A\)'s:
\[ \binom{5}{3}3!. \]Therefore,
\[ 4!\binom{5}{3}3! = 24\cdot10\cdot6 = \boxed{1440}. \]Practice 7
There are two alternating type patterns.
Arrange each group internally:
\[ 2(3!)(3!) = 2\cdot6\cdot6 = \boxed{72}. \]Practice 8
Exactly half of all arrangements have Alex before Blair:
\[ \frac{7!}{2} = \boxed{2520}. \]Practice 9
In seven positions, the unordered position pairs with one person between are
\[ (1,3),\ (2,4),\ (3,5),\ (4,6),\ (5,7). \]There are five pairs.
Alex and Blair can switch positions, and the remaining five people can be arranged freely:
\[ 5\cdot2\cdot5! = 10\cdot120 = \boxed{1200}. \]Practice 10
The count \(6!\) treats the pair as one external block but does not count the two possible internal orders
\[ AB \]and
\[ BA. \]Therefore,
\[ 6!\cdot2 = \boxed{1440}. \]What to Remember
Restricted arrangements are solved by translating conditions into structural counting operations.
Common patterns include:
- fixed positions \(\rightarrow\) place them first;
- forbidden positions \(\rightarrow\) direct counting or complements;
- objects together \(\rightarrow\) block method;
- objects apart \(\rightarrow\) complement or gap method;
- alternating groups \(\rightarrow\) type pattern plus internal arrangements;
- relative order \(\rightarrow\) divide by the number of internal orders;
- overlapping conditions \(\rightarrow\) inclusion–exclusion;
- no fixed points \(\rightarrow\) derangements.
There is no single restriction formula. The central skill is to preserve the exact structure of the legal outcomes while simplifying the count.
Always check whether cases overlap, whether blocks have internal order, whether gaps include the two ends, and whether the restriction is feasible before calculating.
Where We Go Next
Restricted arrangements repeatedly expose the same deeper issue: a raw counting procedure may count one underlying outcome more than once because several descriptions represent the same object.
The next lesson, Symmetry and Overcounting, develops that idea directly and explains how rotations, reflections, identical objects, internal reorderings, and other symmetries create equivalence classes that must be counted only once.