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