Circular Permutations:

Consider the diffenmt arrangements when 4 people sit in a circle. There are 31 arrangements (not 4! As might be expected) as it does not matter where the first object is placed.

circular permutations
In general. n objects can be arranged in a circle in (n-1)! Ways.
• The number of circular permutations of n dissimilar things taken all at a time is (n-1)!
• The number of circular permutations of n dissimilar things in clockwise direction = number of permutations in counterclockwise direction is equal to ½(n-1)!

questions and solutions about circular permutations

Q1. Find the numbers of ways in which 5 men and 5 women can be seated at a round table in such a way that no two persons of same sex sit together.
solution:

If we fix one man round a table
then their permutations =⁴P₄=24
Now if women sit between the two men
then their permutations =⁵P₅=120
So total permutations =24×120=2880

Q2. In how many ways can 4 keys be arranged on a circular key ring?
solution:
Number of keys =4
Fixing one key we have permutation =³P₃=6

Since above figures of arrangement are reflections of each other. Therefore permutations =½×6=3

Q3. How many necklaces can be made from 6 beads of different colours?
solution:
Number of beads =6
Fixing one bead, we have permutation =⁵P₅=120

Since above figures of arrangement are reflections of each other, therefore permutations =½×120=60

Q4. In how many ways 4 boys and 4 girls can be seated on a circular table such that boys and girls sit on alternate positions?

correct: A, solution:
Boys can sit on the circular table in (4-1)! ways. The girls can be seated in 4 places in 4! ways. So. number of ways =3!×4!

Q5. The Governor of the New York calls a meeting of 12 officers. In how many ways can they be seated at a round table?
solution:
Fixing one officer on a particular seat, we have permutations of remaining 11 officers.
=(12-1)!=11!=39916800

Q6. The D.C.Os of 11 districts meet to discuss the law and order situation in their districts. In how many ways can they be seated at a round table, when two particular D.C.Os insist on sitting together?
solution:
Number of D.C.O’s =9
Let D₁and D₂be the two D.C.O’s insisting to sit together so consider them one.
If D₁D₂sit together then permutations

If D₂D₁sit together then permutations

So total permutations =362880+362880=725760
A shortway gives
2!∙(11-2)!=2×362880=725760

Q7. Fatima invites 14 people to a dinner. There are 9 males and 5 females who are seated at two different tables so that guests of one sex seat at one round table and the guests of other sex at the second table. Find the number of ways in which all guests are seated.
solution:
9 males can be seated on a round table =⁸P₈=40320
And 5 females can be seated on a round table =⁴P₄=24
So permutations of both =40320×24=967680.
A shortway gives
=(9-1)!×(5-1)!

Q8. A necklace is to be prepared with 16 beads. 5 out of these beads are similar and 3 of the other beads are similar, How many distinct necklaces can be prepared with the help of these beads?

correct: A, solution:
A necklace of 16 beads can be arranged in 15!/2 ways since a necklace is same when turned upside down. Now it has 5 and 3 similar beads. Hence the total number of distinct necklaces that can be made

The following solution below show that number of clockwise circular permutations is equal to the number of anticlockwise circular permutations

Q9. There are 9 places to be filled with 9 digits from 1 to 9. What is the probability that 6 will get filled before 4?
a ½ b. ¼ c. ⅙ d. 1/10
correct: a, solution:
There is equal probability of 6 getting placed before or after 4. Probability =½.