Circular Permutations Questions and Answers

If we consider the grandparents as a single unit then externally we have 4 people to be arranged in a round table and so externally it is 3! And internally both grandparents are to be arranged so that is 2! So total is 3!x 2! = 12

4 Married couples That is 8 people
Second women can be placed in 6 ways and her husband in 1 way.
Third women can be placed in 4 ways and her husband in 1 way.
Fourth women can be placed in 2 ways and her husband in 1 way.
Total number of ways 6×4×2=48 Ways.

Let the first bead’s position be fixed.

Now we have to arrange the remaining beads = 10 -1 = 9 beads.

These remaining 14 beads can be arranged in ways ¹⁴P₉ = 9!

Also, the necklace has no dependency on the position of the beads, as they can be arranged in clockwise or anti clockwise direction.

Therefore, the required number of ways = \frac{1}{2}\times  (9!)

=181440

We have to calculate total circular permutation in this case when the arrangement of the people is not fixed on a round table.

Total number of people who needs to sit on a table = 5.

If we fix one person around whom these people (5-1) = 4

Number of ways these people can be arranged = 4! = 24

 

Total bangles with Anita = 3 + 5 +2

= 10 bangles.

The number of ways in which they can be arranged = (10-1)!

= 9!

= 362880 ways

Arrangement in the inner circle:

Number of people in the inner circle: 6

When we fix one person, 6-1 = 5!

The remaining 5 children can be arranged in ways = ⁵P₅ = 5!

In the case of circular permutation, as the arrangement is not fixed, the total number of ways students can be arranged = \frac{5!}{2}

Therefore, the arrangement of the inner circle = \frac{5!}{2}= 60

Similarly, in the case of the outer circle,

We fix one student from the group.

Now, there are 11 people remaining who needs to be arranged in a circular formation.

The number of ways these people can be arranged = ¹¹P₁₁ = 11!

As it is circular and nondependent arrangement, the number of ways will be =11!/2

Arrangement = \frac{11\times 10\times 9\times 8\times 7\times 6\times 5\times 4\times 3\times 2\times 1}{2} = 19958400

Therefore, the final number of ways to arrange children in inner as well as outer circles = 19958400+ 60

= 19958460 ways

When A and B needed to be seated together across all the arrangements, these two can be counted as one. so 5 people

Therefore, we now need to arrange five people in a circle = 4! = 24 ways

But A and B can further be seated in two ways which make their arrangement = 2!

Hence, the final arrangement = 24 x 2 = 48 ways

Here, we need to fix the position of three people- Harry and two other people.

We will now treat than as a single unit.

Units that need arrangement = 19

Also, there can be 2 ways in which the other people can sit on either side of the host in 2 ways.

Therefore, the number of ways of their arrangement = 18!

The total number of ways = 18! x 2

Let us assume that 5 girls sit together making them a single unit.

Now we need to arrange only boys and a unit of girls = 10 + 1 = 11 people.

In the case of the circular arrangement, the needed number of ways 10! X 5!

Here we fix the position of one lady and calculate the number of ways people can be seated.

Seating of the ladies = 3!

The number of ways men can be seated = 4!

Therefore, number of ways man and women can be seated alternatively =

= 3! x 4! = 144 ways

If 3 people are sitting at a round table then they can arrange it (3-1)! ways= 2!=2

In the case of a garland, clockwise and anti-clockwise directions will be the same.

Therefore, \frac{(n-1)!}{2}

Here, n=10

Hence, number of ways = \frac{(10-1)!}{2} =\frac{9!}{2}

= 181440

First Select 9 peoples from 20 Peoples = ^{20}\textrm{C}_{9}

Clockwise and anti-clockwise arrangements are different in the case of a seating arrangement around a table.

After Selection Arrangement around a table = (9-1) = 8!

Required number of ways = ^{20}\textrm{C}_{9}\times 8!

Therefore, the total entities become 6

Now, these need to be arranged around a circular round table.

We fix one of these people 6-1 = 5

Therefore, people can be seated in = 5! Ways

Also, the group of Americans can be arranged in 4! Ways.

Hence, the final arrangement will be = 5! x 4!

= 2880

5 Indians can seat around a circular table = 4!

4 American can seat around a circular table = ^{5}\textrm{P}_{4} = 5!

Required Arrangement = 4! x 5!

Number of boys = 5
We consider 4 girls as a single unit.
Therefore, arrangements = 6!
Number of ways girls can be arranged= 4!
Finally, the ways in which these people can be arranged on a circular table = 6! \times 4! =

= 720 x 24
= 17280 ways

In the given case, we can fix the position of one lady.

Therefore, the number of ways in which these ladies can be seated in a circle = (8-1)!

= 7!

= 5040 ways

When P and Q are always sitting together, these two can be considered as a single unit.

The total number of arrangements in this case now will be 3 and 1.

Now, we have to arrange a total 4 people in a circle

Therefore, the number of ways they can be arranged, (4-1)!

= 3!

= 6

Here the two (A and B) can also be interchanged for their places.

Therefore, the total number of ways = 6 x 2

= 12 ways

In such a case, the number of ways can be obtained by removing cases where R and S are sitting together.

The total number of ways is = (5-1)!

= 24

Similarly,

In the case where R and S are always sitting together,

These two can be considered as a single unit.

The total number of arrangements in this case now will be 3 + 1 = 4.

Therefore, the number of ways they can be arranged, (4-1)!

= 3!

= 6

Here the two (R and S) can also be interchanged for their places.

Therefore, the total number of ways = 6 x 2

= 12 ways

Finally, the required number of ways = 24 -12

= 12

The only way to avoid 2 Black bowls placed together is to place them alternatively.

For this, first, we place the three 3 Red bowls alternatively by using (3-1) = 2!

= 2 ways

Now the 3 Black bowls can be placed in the remaining 3 places in 3! = 6 ways

Now we fix the position of the Red bowls the arrangement of the remaining positions will be similar to the linear arrangement.

Hence, the total number of ways = 2! X 3!

= 12

Here we fix the position of one bead and arrange the rest of them.

The number of ways they can be arranged are: \frac{(n-1)!}{2}

Here, n = 10

Therefore, number of ways = \frac{9!}{2}

= \frac{362880}{2}

= 181440

Total number of participants in the competition = 9

Total ranks that need to be awarded = 3

If a person gets the first rank, he cannot get the second rank.

Likewise, if a person secures the second rank, he cannot get the third rank and vice versa.

Therefore, the ways in which the first three ranks can be awarded = 9 x 8 x 7

= 504 ways

 

Total number of bottles = 5 + 4 = 9

Here when we fix the position of one person, we can obtain the number of ways in which these bottles can be arranged.

Therefore, the number of ways in which these can be arranged = 8!

= 40320 ways

Total number of people invited to the party = 5

Members including the host = 5 +1 = 6

The number of ways in which they can be seated on a round table = (6-1)!

One is deducted as we fix the position of one person.

The total number of ways = 5!

= 120 ways

We assume that all the Maths books are kept together.

We count this as a single entity, we have 8 elements or books in total (7 English and 1 Math)

Here we fix the position of one book.

Therefore, the number of ways in which books can be arranged = 7!

Also, the Maths books can further be arranged among themselves= 4! Ways

Therefore, the total number of ways these can be arranged = 7! X 4!

= 5040 x 24

= 120960 ways

The number of possible ways in which boys can sit anywhere in the boat are= ⁸P<sub>6


=</sub> 21060

Seating arrangement for two people P and Q = ⁴P₂

Arrangement for one person R = ⁴P₁

Arrangement for the remaining people = ⁵P₃

Overall arrangement on the boat = ⁴P₂ x ⁴P₁ x ⁵P₃

= 12 x 4 x 20

= 960

Here we fix the position of one ball and arrange the remaining balls in a circle.

Total number of balls that needs to be arranged = 5

The number of ways they can be arranged when one ball is fixed = (5-1)!

= 4!

= 24 ways

In the case of a necklace, the arrangement can be both clockwise as well as anticlockwise.

Total number of beads that needs arrangement = 7

Here we fix the position of the first bead.

Arrangements = (7-1)!

As it is a circular arrangement, the number of ways will be divided by 2

Therefore, the final arrangement of beads = \frac{6!}{2}

= 360 ways

As there are three sets of colors,

Therefore, the number of ways these can be arranged in a bracelet = 2!

Ways to arrange red beads = 3!

Ways to arrange Green beads= 5!

Ways to arrange Black beads = 2!

Ways to arrange all the beads = 2! x 3! x 5! x 2!

= 2 x 6 x 120 x 2

= 2880 ways