Special Arrangements and Restrictions

Scope Label

Core 9758. This note covers the exam-heavy counting cases where a direct use of $n!$, ${}^n{P}_r$, or ${}^n{C}_r$ is not enough.

Use it after Counting Principles, Permutations, and Combinations.

Why Special Arrangements Need Extra Care

Many harder counting questions fail for one of two reasons:

• the same outcome is counted more than once
• a restriction is handled too late

The safest approach is:

1. identify the basic objects
2. identify whether any objects are identical
3. identify the restriction
4. choose a structure that naturally includes the restriction
5. count only after the structure is clear

Arrangements with Identical Objects

If all objects are distinct, $n$ objects can be arranged in $n!$ ways.

If some objects are identical, swapping identical objects does not create a new visible arrangement. Therefore, we divide by the number of invisible swaps.

If there are $n$ objects with identical groups of sizes $n_1,n_2,\dots ,n_k$, the number of distinct arrangements is

$$\frac{n!}{n_1!n_2!\cdots n_k!}.$$

Caption: The division removes arrangements that only differ by swapping identical objects.

Example: MAMMA

The word MAMMA has 5 letters:

• 3 M’s
• 2 A’s

The number of distinct arrangements is

$$\frac{5!}{3!2!}=10.$$

The denominator is not arbitrary. It removes the $3!$ invisible swaps of the M’s and the $2!$ invisible swaps of the A’s.

Restrictions: Main Strategies

Restriction problems should be solved by choosing a counting structure. Do not first count everything and then try random corrections unless the complement method is clearly appropriate.

Caption: Restriction problems are controlled by grouping, slotting, cases, or complement.

Grouping: Objects Must Be Together

If several objects must be together, treat them as a single block first.

Example

Arrange 4 boys and 3 girls in a row if the 3 girls must be together.

Treat the girls as one block:

$$(GGG),B_1,B_2,B_3,B_4.$$

There are 5 units to arrange:

$$5!.$$

Inside the girls’ block, the girls can be arranged in

$$3!$$

ways.

Therefore, the total number is

$$5!\times 3!.$$

The block handles the “together” condition. The internal factorial handles the order within the block.

Slotting: Objects Must Be Separated

If objects must not be adjacent, arrange the other objects first, then place the restricted objects in available gaps.

Example

Arrange 5 boys and 3 girls in a row if no two girls are adjacent.

First arrange the boys:

$$5!.$$

This creates 6 slots:

$$\_B\_B\_B\_B\_B\_$$

Choose 3 of the 6 slots for the girls and arrange the girls:

$${}^6{C}_3\times 3!.$$

So the total number is

$$5!\times {}^6{C}_3\times 3!.$$

This method works because each selected slot can contain at most one girl.

Cases

Use cases when the problem naturally splits into non-overlapping possibilities.

The cases must be:

• mutually exclusive: no outcome appears in two cases
• exhaustive: all valid outcomes are covered

Example

Choose 3 students from 5 boys and 4 girls with at least 2 boys.

Split into cases:

• exactly 2 boys and 1 girl
• exactly 3 boys

So the count is

$${}^5{C}_2{}^4{C}_1+{}^5{C}_3.$$

Complement Method

Use complement when counting the unwanted outcomes is easier than counting the wanted outcomes directly.

The structure is:

$$\text{wanted}=\text{total}-\text{unwanted}.$$

Example

From 8 students, choose a committee of 4 with at least one girl. Suppose there are 5 boys and 3 girls.

Total committees:

$${}^8{C}_4.$$

Unwanted committees with no girls:

$${}^5{C}_4.$$

Therefore, the number of valid committees is

$${}^8{C}_4-{}^5{C}_4.$$

Circular Permutations

In a row, positions are distinguishable. In an unnumbered circle, rotations are usually equivalent.

For $n$ distinct objects around an unnumbered circle, the number of arrangements is

$$(n-1)!.$$

Reason: fix one object to remove rotational freedom, then arrange the remaining $n-1$ objects.

Caption: Rotating the whole circular arrangement does not produce a new arrangement.

Numbered versus unnumbered circular seats

If the seats are numbered, the positions are distinguishable. Then arranging $n$ people in $n$ numbered seats gives

$$n!$$

ways.

If the seats are unnumbered around a round table, rotations are equivalent. Then the count is

$$(n-1)!.$$

Reflections

In most H2 seating questions, clockwise and anticlockwise orders are different unless the question explicitly says arrangements that are mirror images are the same.

Do not divide by $2$ unless the problem says reflections are equivalent.

Circular Restrictions

Circular restriction problems often combine the circular idea with grouping or slotting.

Objects together in a circle

If 3 friends must sit together among 7 people around an unnumbered round table:

1. treat the 3 friends as one block
2. arrange the block plus the other 4 people around a circle
3. arrange the 3 friends inside the block

There are 5 circular units, so

$$(5-1)!\times 3!.$$

Objects separated in a circle

If no two of a certain group may sit together, first arrange the unrestricted group around the circle, then place the restricted group into gaps.

The exact count depends on whether the gaps are distinguishable after the first circular arrangement is fixed, and on whether the restricted objects are distinct.

The main warning is:

Do not use row-slotting blindly in a circle.

Mixed Counting Problems

Many exam questions mix several ideas. A useful workflow is:

1. translate the wording into a precise outcome
2. decide whether order matters
3. decide whether objects are distinct or identical
4. decide whether the arrangement is linear or circular
5. identify restrictions
6. decide whether grouping, slotting, cases, or complement is most natural
7. write the count in stages before evaluating

Worked Core Examples

Example 1: Identical objects with restrictions

How many arrangements of BANANA begin and end with A?

BANANA has 6 letters: A, A, A, N, N, B.

Fix A at both ends:

$$A\_\_\_\_A.$$

The middle four positions contain A, N, N, B.

So the number of arrangements is

$$\frac{4!}{2!}=12.$$

Example 2: Not all together

Arrange 5 distinct people in a row if 3 specified people are not all together.

Use complement:

$$\text{not all together}=\text{total}-\text{all together}.$$

Total:

$$5!.$$

All together: treat the 3 specified people as a block. Then there are 3 units to arrange, and the block has $3!$ internal arrangements:

$$3!\times 3!.$$

So the answer is

$$5!-3!\times 3!.$$

This is not the same as “all separated”.

Example 3: Committee with restrictions

From 6 boys and 5 girls, choose a committee of 4 with at least 2 girls.

Use cases:

$${}^5{C}_2{}^6{C}_2+{}^5{C}_3{}^6{C}_1+{}^5{C}_4.$$

The cases correspond to exactly 2 girls, exactly 3 girls, and exactly 4 girls.

Common Pitfalls

• Treating “not all together” as “all separated”.
• Forgetting to arrange objects inside a grouped block.
• Creating slots before arranging the unrestricted objects.
• Forgetting to check whether there are enough slots.
• Dividing by $2$ for circular reflections when the question does not make reflections equivalent.
• Using row formulas directly in circular arrangements.
• Splitting into cases that overlap.
• Using complement but subtracting the wrong unwanted set.

Links

• Permutations and Combinations
• Counting Principles, Permutations, and Combinations