VIT - VITEEE 2025
National level exam conducted by VIT University, Vellore | Ranked #11 by NIRF for Engg. | NAAC A++ Accredited | Last Date to Apply: 31st March | NO Further Extensions!
CIRCULAR PERMUTATIONS, DIFFERENT CASES OF GEOMETRICAL ARRANGEMENTS is considered one of the most asked concept.
72 Questions around this concept.
The number of ways in which 6 men and 5 women can dine at a round table if no two women are to sit together is given by
Find the number of ways in which seating arrangement can be made around a circular table of 8 members.
Find the number of ways of arranging 12 flowers in a garland.
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Let the difference of no. of lines and no. of triangles that can be formed by 7 points (no 3 of them are collinear) and 4 collinear points be a then $|a|$ is.
Let the difference of no. of lines and no. of triangles that can be formed by 8 non-collinear points (No 3 of them are collinear) be a then |a| is?
12 persons are to be arranged at a round table. If two particular persons among them are not to be side by side, what is the total number of arrangements?
The number of ways in which 5 boys and 3 girls can be seated on a round table if a particular boy $\mathrm{B}_1$ and a particular girl $\mathrm{G}_1$ never sit adjacent to each other, is :
National level exam conducted by VIT University, Vellore | Ranked #11 by NIRF for Engg. | NAAC A++ Accredited | Last Date to Apply: 31st March | NO Further Extensions!
Ranked #42 among Engineering colleges in India by NIRF | Highest CTC 50 LPA , 100% Placements | Last Date to Apply: 28th March
In the conference hall, the Manager calls a meeting. Find How many ways five other employees can be arranged in front of the manager if the chair is arranged in a circular order and the position of Manager is already fixed.
What is the number of circular permutations of 7 distinct objects out of 12 distinct objects in a necklace?
What is the number of circular permutations of 7 objects out of 12 available objects?
Rule for Geometrical Permutations -
There are n points in the plane such that no three of them are on the same straight line except m of them which are on same straight line, then the number of straight lines formed by joining them is/are ${ }^n c_2-{ }^m c_2+1$ and number triangles is/are ${ }^n c_3-{ }^m c_3$.
Let's say there is a round table with 6 chairs all identical and 6 persons have to sit. For the first person there is only one choice to make as all chairs are identical, so wherever he may sit doesn't matter. Now when $1^{\text {st }}$ person has sit, $2^{\text {nd }}$ person with respect to $1^{\text {st }}$ have five choices to sit, directly opposite or left or $2^{\text {nd }}$ from left or right or $2^{\text {nd }}$ from right to $1^{\text {st }}$ person, in the same ways $3^{\text {rd }}$ person will have 4 choices, $4^{\text {th }}$ person will have 3 choice, $5^{\text {th }}$ person will have 2 choices, and last person 1 choice, so in that way total $5 \times 4 \times 3 \times 2 \times 1=(6-1)$ ! permutations are possible.
We can generalize this result as an object that can be arranged along with a circular table in (n-1)! Ways
Example: In how many ways 7 people can be arranged along a circular table having 7 identical chairs.
Solution: using the above concept, it can be done in (7-1)! = $6!\mid$
Necklace, Garland Type Questions
If clockwise and anticlockwise permutations are same in a circular permutation, (as in case of garlands or necklace formation), the number of permutations becomes $(1 / 2)(n-1)$ !
Since as in the case of necklace and garland if we flip the bead or garland then anticlockwise arrangements become clockwise but they are identical because the objects are identical, hence two arrangements are reduced to one causing the total number of permutations to be halved.
Example: Find the ways in which 10 different beads can be arranged to form a necklace?
Solution: Using circular permutations the total number of permutations $=(10-1)$ !
Now since clockwise and anticlockwise arrangements give the same permutation so the total number of permutations becomes $(1 / 2)(9!)$
If clockwise and anticlockwise permutations are same in a circular permutation, (as in case of garlands or necklace formation), the number of permutations becomes (1/2)(n-1)!.
Since as in the case of necklace and garland if we flip the bead or garland then anticlockwise arrangements become clockwise but they are identical because the objects are identical, hence two arrangements are reduced to one causing the total number of permutations to be halved.
Example: Find the ways in which 10 different beads can be arranged to form a necklace?
Solution: Using circular permutations the total number of permutations $=(10-1)$ !
Now since clockwise and anticlockwise arrangements give the same permutation so the total number of permutations becomes (1/2)(9!)
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