Circular permutations are an exciting part of combinatorics, where we arrange objects in a circular pattern, like seating people around a table or placing keys on a ring. Unlike linear arrangements, where the starting position matters, circular arrangements treat rotations of the same arrangement as identical. This blog will dive into what circular permutations are, their key formula, strategies to solve common problems, and a variety of practice questions for both linear and circular arrangements, complete with detailed, step-by-step solutions to help you master the topic.

What is Circular Permutation?

Circular permutation is the number of ways to arrange $n$ distinct objects in a circle, where rotating the arrangement does not create a new one. For example, if three friends, A, B, and C, sit around a table, the seating orders ABC, BCA, and CAB are considered the same because they can be obtained by rotating the table. This rotational symmetry is what makes circular permutations unique compared to linear permutations, where ABC and BCA would be distinct arrangements.

Formula for Circular Permutation

For $n$ distinct objects arranged in a circle, the number of distinct arrangements is:

$$(n–1)!$$

Why $(n–1)!$ ? In a circular arrangement, all rotations of a single arrangement are identical. To avoid counting these rotations, we fix one object in a position (say, at the “top” of the circle) and arrange the remaining $n–1$ objects around it. This gives us $(n–1)!$ arrangements. For example, withcrets3 objects (A, B, C), fixing A and arranging B and C gives $2!=2$ arrangements: A-B-C and A-C-B.

Special Case: Arrangements with Reflection (e.g., Keys on a Ring)
If the circle can be flipped, like a keychain where flipping it over makes the arrangement look the same (e.g., A-B-C is the same as A-C-B when flipped), we divide by 2 to account for reflections:

$$\frac{(n–1)!}{2}$$

Linear Permutation Formula for Comparison
Arranging $n$ distinct objects in a straight line gives:

$$n!$$

This is because linear arrangements care about the exact position of each object, with no rotational symmetry. For 3 objects, linear arrangements are $3!=6$ (ABC, ACB, BAC, BCA, CAB, CBA), while circular arrangements are $(3–1)!=2$.

Key Strategies for Solving Permutation Questions

To make solving permutation problems easier, especially when specific conditions are given, follow these strategies, which are particularly helpful when certain keywords appear in the question:

1. Always Together/Not Separated: When certain objects must stay together, treat them as a single unit. Arrange this unit along with the other objects, then arrange the objects within the unit. For example, if two friends, A and B, must sit together in a row, we group them as (AB). For 3 people (A, B, C), we have [(AB), C]. Arrange the 2 units: $2!=2$ ways ((AB)C or C(AB)). Within (AB), A and B can swap: $2!=2$ ways (AB or BA). Total = $2!\times 2!=4$. This reduces the problem’s complexity by treating the group as one entity.
2. All Never Together/All Do Not Come Together: Calculate the total number of arrangements without restrictions, then subtract the cases where the specified objects are together. For example, for 3 people (A, B, C) in a row where A and B are never together, total arrangements = $3!=6$. A and B together: treat (AB) as one unit, $2!\times 2!=4$. Subtract: $6–4=2$ arrangements where A and B are not together.
3. No Two Objects Seated Together: To ensure no two objects (e.g., girls) are adjacent, arrange the other group (e.g., boys) first, creating gaps. Place the restricted group in these gaps to prevent adjacency. For example, with 3 boys and 3 girls in a row, with no two girls together, arrange boys: _B1_B2_B3_. This creates 4 gaps (before B1, between B1 and B2, B2 and B3, after B3). Choose 3 gaps for girls: $C_3^4=4$. Arrange 3 girls: $3!=6$. Arrange 3 boys: $3!=6$. Total = $4\times 6\times 6=144$.
4. Alternating Arrangements: For alternating arrangements (e.g., boy-girl-boy-girl), fix a pattern (e.g., BGBG or GBGB), arrange each group in their designated positions, and account for both patterns if needed. For 2 boys and 2 girls in a row, for BGBG, arrange 2 boys in boy positions: $2!=2$, and 2 girls in girl positions: $2!=2$. Total for BGBG = $2!\times 2!=4$. Repeat for GBGB = 4. Total = $4+4=8$.

Practice Questions: Linear Arrangements

Let’s dive into the practice questions, solving them step-by-step with detailed explanations to ensure you understand every part of the process.

Question 1: 4 Boys and 4 Girls Seated in a Row

a) No Restrictions

Problem: Find the number of ways to arrange 4 boys and 4 girls in a row with no restrictions.

Solution:
– We have 4 boys (B1, B2, B3, B4) and 4 girls (G1, G2, G3, G4), making a total of 8 distinct people.
– Since it’s a linear arrangement with no restrictions, we arrange all 8 people in a row.
– The number of ways to arrange 8 distinct objects is:

$$8!$$

– Calculate: $8!=8\times 7\times 6\times 5\times 4\times 3\times 2\times 1=40320$.
– This counts all possible seating arrangements, such as B1-G1-B2-G2-B3-G3-B4-G4, G1-B1-G2-B2-G3-B3-G4-B4, and so on.

Answer: 40320

b) All Boys Together and All Girls Together

Problem: Find the number of arrangements where all boys are together and all girls are together.

Solution:
– We need all 4 boys to sit together as a block and all 4 girls to sit together as another block, e.g., (B1B2B3B4)(G1G2G3G4) or (G1G2G3G4)(B1B2B3B4).
– Treat the group of boys as one unit (Boys) and the group of girls as one unit (Girls). Now we have 2 units: [(Boys), (Girls)].
– Arrange these 2 units in a row: $2!=2$ ways (Boys-Girls or Girls-Boys).
– Within the Boys unit, the 4 boys can be arranged:

$$4!=4\times 3\times 2\times 1=24$$

– Within the Girls unit, the 4 girls can be arranged:

$$4!=24$$

– Total arrangements:

$$2!\times 4!\times 4!=2\times 24\times 24=1152$$

– Example arrangements include B1-B2-B3-B4-G1-G2-G3-G4 or G1-G2-G3-G4-B1-B2-B3-B4, with internal rearrangements of boys and girls.

Answer: 1152

c) All Girls Are Together

Problem: Find the number of arrangements where all girls are together.

Solution:
– We need all 4 girls to sit together as a single block, e.g., B1-B2-(G1G2G3G4)-B3-B4.
– Treat the 4 girls as one unit: (Girls). Now we have 5 units: [(Girls), B1, B2, B3, B4].
– Arrange these 5 units in a row:

$$5!=5\times 4\times 3\times 2\times 1=120$$

– Within the Girls unit, arrange the 4 girls:

$$4!=24$$

– Total arrangements:

$$5!\times 4!=120\times 24=2880$$

– Example arrangement: B1-B2-(G1-G2-G3-G4)-B3-B4, where the girls’ block can be rearranged (e.g., G2-G1-G4-G3) and the 5 units can be placed in different orders.

Answer: 2880

d) All Girls Are Never Together

Problem: Find the number of arrangements where all girls are not together.

Solution:
– To find arrangements where the 4 girls are not all together, we subtract the cases where they are together from the total arrangements.
– Total arrangements (no restrictions, from part a):

$$8!=40320$$

– Girls together (from part c):

$$5!\times 4!=2880$$

– Arrangements where girls are not all together:

$$8!–(5!\times 4!)=40320–2880=37440$$

– This includes arrangements where girls are scattered or partially grouped, e.g., G1-B1-G2-B2-G3-B3-G4-B4, ensuring the 4 girls do not form a single block.

Answer: 37440

e) No Two Girls Are Seated Together

Problem: Find the number of arrangements where no two girls are seated together.

Solution:
– To ensure no two girls are adjacent, we first arrange the 4 boys to create gaps where girls can be placed.
– Arrange the 4 boys in a row: _B1_B2_B3_B4_. This creates 5 gaps: before B1, between B1 and B2, B2 and B3, B3 and B4, and after B4.
– Choose 4 out of 5 gaps for the 4 girls to prevent adjacency:

$$C_4^5=5$$

– Arrange the 4 girls in these 4 gaps:

$$4!=4\times 3\times 2\times 1=24$$

– Arrange the 4 boys in their positions:

$$4!=24$$

– Total arrangements:

$$C_4^5\times 4!\times 4!=5\times 24\times 24=2880$$

– Alternatively, treat the gaps as positions for girls:

$$P_4^5=5\times 4\times 3\times 2=120$$

– Multiply by boys’ arrangements:

$$120\times 4!=120\times 24=2880$$

– Example arrangement: G1-B1-G2-B2-G3-B3-G4-B4, where girls are separated by boys.

Answer: 2880

f) Boys and Girls Seated Alternately

Problem: Find the number of arrangements where boys and girls are seated alternately.

Solution:
– We need boys and girls to alternate, e.g., B-G-B-G-B-G-B-G or G-B-G-B-G-B-G-B.
– Case 1: B-G-B-G-B-G-B-G pattern
– Arrange 4 boys in the boy positions:

$$4!=24$$

– Arrange 4 girls in the girl positions:

$$4!=24$$

– Total for this pattern:

$$4!\times 4!=24\times 24=576$$

– Case 2: G-B-G-B-G-B-G-B pattern
– Similarly:

$$4!\times 4!=576$$

– Total arrangements:

$$576+576=1152$$

– Example arrangements: B1-G1-B2-G2-B3-G3-B4-G4 or G1-B1-G2-B2-G3-B3-G4-B4, with internal rearrangements.

Answer: 1152

Question 2: There are 4 girls and 3 boys, among whom 2 are brothers then find:

a) Two Brothers Seated Together

Problem: Find the number of arrangements where two brothers are seated together, given 4 girls and 3 boys (2 brothers + 1 other boy).

Solution:
– We have 4 girls (G1, G2, G3, G4), 2 brothers (B1, B2), and 1 other boy (B3), total 7 people.
– Treat the two brothers as one unit: (B1B2). Now we have 6 units: [(B1B2), G1, G2, G3, G4, B3].
– Arrange these 6 units in a row:

$$6!=6\times 5\times 4\times 3\times 2\times 1=720$$

– Within (B1B2), the brothers can swap positions:

$$2!=2$$

– Total arrangements:

$$6!\times 2!=720\times 2=1440$$

– Example arrangement: B3-G1-G2-(B1B2)-G3-G4, where (B1B2) could be B2B1, and the 6 units can be rearranged.

Answer: 1440

b) Brothers Are Not Together

Problem: Find the number of arrangements where the brothers are not together.

Solution:
– Total people: 4 girls + 2 brothers + 1 boy = 7.
– Total arrangements (no restrictions):

$$7!=7\times 6\times 5\times 4\times 3\times 2\times 1=5040$$

– Brothers together (from part a):

$$6!\times 2!=1440$$

– Brothers not together:

$$7!–(6!\times 2!)=5040–1440=3600$$

– Example arrangement: G1-B1-G2-B3-G3-B2-G4, where B1 and B2 are separated by at least one person.

Answer: 3600

c) Exactly One Person Between Two Brothers

Problem: Find the number of arrangements with exactly one person between the brothers.

Solution:
– We need a pattern like (B1-P-B2), where P is any one person (girl or boy) between the brothers.
– Choose the person P from the remaining 5 people (4 girls + 1 boy):

$$C_1^5=5$$

– Treat (B1-P-B2) as one unit. Now we have 5 units: [(B1-P-B2), G1, G2, G3, B3].
– Arrange these 5 units:

$$5!=5\times 4\times 3\times 2\times 1=120$$

– Within (B1-P-B2), the brothers can swap:

$$2!=2$$

– Total arrangements:

$$C_1^5\times 5!\times 2!=5\times 120\times 2=1200$$

– Example arrangement: G1-(B1-G2-B2)-G3-B3-G4, where G2 is between B1 and B2, and the 5 units are rearranged.

Answer: 1200

d) Sister Always Between Brothers

Problem: Find the number of arrangements in which the sister is always between the brothers.

Solution: – Form a unit (B1-S-B2), where S is the sister.
No need to choose the sister from the 4 girls (because the sister already knows who her brothers are 😄)

– Treat (B1-S-B2) as one unit. Now we have 5 units: [(B1-S-B2), G1, G2, G3, B3].
– Arrange these 5 units:

$$5!=120$$

– Within (B1-S-B2), the brothers can swap:

$$2!=2$$

– Total arrangements:

$$5!\times 2!=120\times 2=240$$

– Example arrangement: G1-(B1-G2-B2)-G3-B3-G4, where G2 is the sister, and units are rearranged.

Answer: 960

Practice Questions: Circular Arrangements

Question 1: 4 Boys and 4 Girls at a Circular Table

a) No Restrictions

Problem: Find the number of ways to arrange 4 boys and 4 girls around a circular table.

Solution:
– Total people: 4 boys + 4 girls = 8.
– In a circular arrangement, rotations are considered the same. For $n$ distinct objects, the number of circular arrangements is:

$$(n–1)!$$

– Here, $n=8$, so:

$$(8–1)!=7!=7\times 6\times 5\times 4\times 3\times 2\times 1=5040$$

– This counts arrangements like B1-G1-B2-G2-B3-G3-B4-G4, where rotating the table (e.g., G1-B2-G2-B3-G3-B4-G4-B1) is the same.

Answer: 5040

b) All Boys Are Together

Problem: Find the number of arrangements where all boys are together.

Solution:
– Treat the 4 boys as one unit: (B1B2B3B4). Now we have 5 units: [(Boys), G1, G2, G3, G4].
– Arrange these 5 units in a circle:

$$(5–1)!=4!=4\times 3\times 2\times 1=24$$

– Arrange the 4 boys within the unit:

$$4!=24$$

– Total arrangements:

$$4!\times 4!=24\times 24=576$$

– Example arrangement: (B1-B2-B3-B4)-G1-G2-G3-G4, where the boys’ block and girls can be rearranged, and rotations are equivalent.

Answer: 576

c) No Two Boys Are Together

Problem: Find the number of arrangements where no two boys are together.

Solution:
– Arrange the 4 girls in a circle:

$$(4–1)!=3!=3\times 2\times 1=6$$

– In a circular arrangement, 4 girls create 4 gaps (one between each pair of girls).
– Place the 4 boys in these 4 gaps, one per gap, to ensure no two boys are adjacent:

$$4!=24$$

– Total arrangements:

$$3!\times 4!=6\times 24=144$$

– Example arrangement: G1-B1-G2-B2-G3-B3-G4-B4, where boys are separated by girls, and rotations are equivalent.

Answer: 144

Question 2: 3 Sisters and 8 Other Girls in a Circular Order

Problem: Find the number of ways to arrange 11 girls (3 sisters + 8 others) in a circular order so that the three sisters are not seated together.

Solution:
– Total girls: 3 sisters + 8 others = 11.
– Total circular arrangements:

$$(11–1)!=10!=3628800$$

– Sisters together: Treat the 3 sisters as one unit. Now we have 9 units (1 sisters unit + 8 others). Arrange in a circle:

$$(9–1)!=8!=40320$$

– Arrange the 3 sisters within the unit:

$$3!=6$$

– Total for sisters together:

$$8!\times 3!=40320\times 6=241920$$

– Sisters not together:

$$10!–(8!\times 3!)=3628800–241920=3386880$$

– Example: An arrangement where sisters are separated, like S1-G1-S2-G2-S3-G3-G4-G5-G6-G7-G8, with rotations equivalent.

Answer: 3386880

Question 3: 11 Committee Members with President and Secretary Together

Problem: Find the number of arrangements where 11 members sit at a round table with the President and Secretary together.

Solution:
– Treat the President (P) and Secretary (S) as one unit: (PS). They can swap within the unit:

$$2!=2$$

– Now we have 10 units: [(PS), M1, M2, …, M9]. Arrange these in a circle:

$$(10–1)!=9!=9\times 8\times 7\times 6\times 5\times 4\times 3\times 2\times 1=362880$$

– Total arrangements:

$$9!\times 2!=362880\times 2=725760$$

– Example arrangement: (P-S)-M1-M2-M3-M4-M5-M6-M7-M8-M9, with rotations and P-S or S-P swaps.

Answer: 725760

Question 4: 5 Keys on a Ring with Rotations and Reflections

Problem: Find the number of distinct ways to arrange 5 keys on a ring, where arrangements are the same if rotated or flipped.

Solution:
– For a ring (like a keychain), arrangements are identical under rotations and reflections (flipping the ring).
– The formula for distinct arrangements of $n$ distinct objects on a ring with rotations and reflections is:

$$\frac{(n–1)!}{2}$$

– For 5 keys: $n=5$. Calculate:

$$(5–1)!=4!=4\times 3\times 2\times 1=24$$

– Then:

$$\frac{24}{2}=12$$

– Example: For keys A, B, C, D, E, an arrangement like A-B-C-D-E clockwise is the same as A-E-D-C-B counterclockwise (its reflection). Thus, we count them as one.

Answer: 12

Additional Practice Questions

Question 5: Linear Arrangement with Specific Pairing

Problem: 5 boys and 3 girls are seated in a row. Find the number of arrangements where a specific boy (B1) and a specific girl (G1) are seated together.

Solution:
– Treat B1 and G1 as one unit: (B1G1). Now we have 7 units: [(B1G1), B2, B3, B4, B5, G2, G3].
– Arrange these 7 units in a row:

$$7!=7\times 6\times 5\times 4\times 3\times 2\times 1=5040$$

– Within (B1G1), they can swap:

$$2!=2$$

– Total arrangements:

$$7!\times 2!=5040\times 2=10080$$

– Example arrangement: B2-(B1-G1)-B3-G2-B4-G3-B5, with B1 and G1 together and units rearranged.

Answer: 10080

Question 6: Circular Arrangement with No Two Girls Together

Problem: 5 boys and 3 girls are seated around a circular table. Find the number of arrangements where no two girls are together.

Solution:
– Arrange the 5 boys in a circle:

$$(5–1)!=4!=4\times 3\times 2\times 1=24$$

– This creates 5 gaps between the boys (since it’s a circle).
– Choose 3 of these 5 gaps for the 3 girls to ensure no two girls are adjacent:

$$C_3^5=\frac{5\times 4\times 3}{3\times 2\times 1}=10$$

– Arrange the 3 girls in these 3 gaps:

$$3!=6$$

– Total arrangements:

$$4!\times C_3^5\times 3!=24\times 10\times 6=1440$$

– Example arrangement: B1-G1-B2-G2-B3-G3-B4-B5, with girls separated by boys, rotations equivalent.

Answer: 1440

Question 7: Circular Arrangement with Two Groups Together

Problem: 6 people (3 men, 3 women) are seated around a table. Find the number of arrangements where all men are together and all women are together.

Solution:
– Treat the 3 men as one unit (Men) and the 3 women as one unit (Women): [(Men), (Women)].
– Arrange these 2 units in a circle:

$$(2–1)!=1!=1$$

– Arrange the 3 men within their unit:

$$3!=6$$

– Arrange the 3 women within their unit:

$$3!=6$$

– Total arrangements:

$$1\times 3!\times 3!=1\times 6\times 6=36$$

– Example arrangement: (M1-M2-M3)-(W1-W2-W3), with internal rearrangements and rotations equivalent.

Answer: 36

Conclusion

Circular permutations bring a unique perspective to combinatorics by accounting for rotational symmetry, distinguishing them from linear arrangements. By mastering the formula $(n–1)!$, along with strategies like grouping objects, subtracting restricted cases, or arranging alternately, you can confidently solve a wide range of problems. The practice questions above, with their detailed solutions, should help you build a strong foundation. Keep practicing, and explore more combinatorics topics to enhance your skills!