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MHT-CET Maths · Permutations and Combinations

Circular Arrangements

n distinct people around a round table sit in (n − 1)! ways because rotations are the same seating — and every row move (block, gap, complement) carries over with that one change.

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Why this matters

9 PYQs at 56% HARD — level with Selections as the densest HARD page in the chapter, and all nine are from 2023 onwards. The stems are a special pair or trio that must not sit together, girls kept apart around a table, two tables of different sizes, alternating genders with a couple kept together, and a hat-colouring of a five-seat circle. Each is a row problem with (n − 1)! in place of n!, plus the gap method, which is where most of the marks are lost.

Concept 1 of 3: (n − 1)! Around a Table, and Girls Apart via the Gaps Between Boys

Fix one person to kill the rotations; the other n−1n - 1 arrange in (n−1)!(n-1)! ways. To keep girls apart, seat the boys first — nn boys around a table make nn gaps (not n+1n + 1) — and drop the girls into different gaps.

Definition

  • nn distinct people, round table: (n−1)!(n - 1)!. If clockwise and anticlockwise count as the same (a necklace), (n−1)!2\dfrac{(n-1)!}{2}.
  • 66 boys and 55 girls, no two girls together: boys 5!=1205! = 120; 66 gaps; girls 6P5=720{}^6P_5 = 720; total 86,40086{,}400.
  • Feasibility: gg girls need g≤bg \le b gaps around a circle (against g≤b+1g \le b + 1 in a row).
  • Two tables of 1212 and 99 for 2121 friends: choose the first table's people 21C12{}^{21}C_{12}, then seat both tables 11!11! and 8!8!: 21!12! 9!⋅11!⋅8!=21!12⋅9=359⋅19!\dfrac{21!}{12!\,9!}\cdot 11!\cdot 8! = \dfrac{21!}{12 \cdot 9} = \dfrac{35}{9}\cdot 19!.

Circular permutations

(n−1)!girls apart: (b−1)!×bPg(n - 1)! \qquad \text{girls apart: } (b - 1)! \times {}^{b}P_{g}

Worked example

In how many ways can 44 boys and 33 girls sit around a round table so that no two girls are adjacent?
Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 19 April Shift I · Q102Moderate

Example 1 · Permutations and Combinations · Circular Arrangements

The number of ways, in which 6 boys and 5 girls can sit at a round table, if no two girls are to sit together, is

Using n + 1 gaps on a circle

Six boys in a ROW make 77 gaps; around a TABLE they make 66. 7P5{}^7P_5 instead of 6P5{}^6P_5 is how the wrong options are built.
Drill 1 more on (n − 1)! around a table, and girls apart via the gaps between boys

Concept 2 of 3: Never Together Around a Table: Total Minus the Glued Block

A group that must not sit together is counted by complement: all seatings (n−1)!(n-1)! minus the seatings with the group glued, which is (n−k)!(n - k)! for the blocks times k!k! inside. For just two people, the gap method is equally quick.

Definition

  • 66 boys + 44 girls, two special boys and a special girl never all together: total 9!9!; glued trio: (10−3+1−1)!×3!=7!×6=30,240(10 - 3 + 1 - 1)! \times 3! = 7! \times 6 = 30{,}240; answer 362,880−30,240=332,640362{,}880 - 30{,}240 = 332{,}640.
  • Blocks around a circle: mm units seat in (m−1)!(m - 1)! ways, so a block of kk among nn people gives (n−k)!⋅k!(n - k)!\cdot k! glued seatings.
  • 55 boys + 33 girls, B1B_1 and G1G_1 never adjacent, by gaps: seat the other 66 in 5!5!; 66 gaps; B1,G1B_1, G_1 into two different gaps 6P2=30{}^6P_2 = 30: 120×30=3600=5×6!120 \times 30 = 3600 = 5 \times 6!.
  • 'Never all three together' (trio) differs from 'no two of the three adjacent'; the PYQ wording is the trio, so only the fully-glued case is subtracted.

Glued block on a circle

(n−1)!−(n−k)! k!(n - 1)! - (n - k)!\,k!

Worked example

Seven people sit around a round table. In how many ways can they sit if two particular people are not adjacent?
Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 26 April Shift II · Q124Hard

Example 2 · Permutations and Combinations · Circular Arrangements

The number of ways in which 6 boys and 4 girls can be seated around a round table such that 2 special boys and a special girl never sit together is

Gluing on a circle with row arithmetic

A block among 1010 people on a circle leaves 88 units, seated in 7!7! ways, not 8!8!. Using 8!×3!8! \times 3! subtracts too much and lands on a non-option.
Drill 1 more on never together around a table: total minus the glued block

Concept 3 of 3: Alternating Seats With a Couple Together, and Colouring a Circle

Two 2025 stems that look exotic are ordinary once the frame is fixed. Alternating genders fixes which seats are male and which female, so the count is (boys)!×(girls)!\text{(boys)}! \times \text{(girls)}! once the couple's block is placed. Colouring seats so neighbours differ is a small recurrence.

Definition

  • Mother, father and 44 boys + 44 girls, alternating, parents together: the parents form the one male–female adjacent pair; fixing that block fixes the alternation around the table. Remaining boys 4!4! and girls 4!4!: 576576, the official answer. (Counting the block's two orientations separately would double it; the key keeps one.)
  • Proper colourings of a cycle of nn seats with kk colours, neighbours different: (k−1)n+(−1)n(k−1)(k-1)^n + (-1)^n(k-1). Five seats, three colours: 25−2=302^5 - 2 = 30.
  • Why: a PATH of nn seats has k(k−1)n−1k(k-1)^{n-1} colourings; closing the path into a cycle subtracts those where the two ends match, which gives the recurrence above.

Cycle colourings

P(Cn,k)=(k−1)n+(−1)n(k−1)P(C5,3)=32−2=30P(C_n, k) = (k-1)^n + (-1)^n (k-1) \qquad P(C_5, 3) = 32 - 2 = 30

Worked example

Four chairs are placed around a round table. Each chair is painted one of 33 colours so that no two adjacent chairs share a colour. How many colourings?
Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 4th May Shift 2 · Q136Hard

Example 3 · Permutations and Combinations · Circular Arrangements

Five persons A, B, C, D and E are seated in a circular arrangement, if each of them is given a hat of one of the three colours red, blue and green, then the number of ways of distributing the hats such that the person seated in adjacent seats get different coloured hats, is

Treating the parents' block as a row block

In the alternating stem the parents' block has no free 2!2! — the alternation decides which side the father sits. The official answer is 4!×4!=5764! \times 4! = 576; 11521152 is the doubled count.
Drill 1 more on alternating seats with a couple together, and colouring a circle

Summary — formulas & gotchas at a glance

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