Maths > Permutations and Combinations > 4.0 Circular Permutations
Permutations and Combinations
1.0 Factorial notation
2.0 Basic Principle of Counting
3.0 Permutation
4.0 Circular Permutations
5.0 Combinations
6.0 Restricted Selection
7.0 Restricted Arrangement
8.0 Greatest Term
9.0 Possible selections from $n$ distinct objects
10.0 Possible Selection from $n$ Identical Objects
11.0 Possible Selection from $n$ Objects having Distinct and Identical Objects
12.0 Total Number of Possible Divisors for a Given Natural Number
13.0 Sum of all Possible Divisors of a Natural Number
14.0 Exponent of a Prime Number in $n!$
15.0 Division and Distribution of Objects
16.0 Multinomial Theorem
4.2 Permutation when taken $r$ at a time
When clockwise and anti-clockwise are different, $$={{^n{P_r}} \over r}$$
When clockwise and anti-clockwise are same, $$={{^n{P_r}} \over 2r}$$
Question 11. There are $10$ distinct color carriages in a Ferris Wheel. In how many ways can these be arranged? What is the maximum number of arrangements possible if only $8$ out of these are used to build it?
Solution: For a ferris wheel, clockwise and anticlockwise are not distinguishable.
Thus total arrangements is, $$\begin{equation} \begin{aligned} = \frac{{(10 - 1)!}}{2} \\ = \frac{{9!}}{2} \\ = 181440 \\\end{aligned} \end{equation} $$
Now, when only $8$ out of them are selected.
Then total arrangements is given by, $$\begin{equation} \begin{aligned} = \frac{{{}^{10}{P_8}}}{{2(8)}} \\ = \frac{{10!}}{{16(10 - 8)!}} \\ = 1,13,400 \\\end{aligned} \end{equation} $$
Question 12. A carousel has $12$ horses.
(A) Find the number of arrangements possible if every horse is occupied exactly by one kid.
(B) Find the number of arrangements possible if only $6$ horses are to be occupied when the number of horses is decreased to $8$.
Solution:
(A) Here the clockwise and anticlockwise are different.
Hence total arrangements is given by, $$\begin{equation} \begin{aligned} = (12 - 1)! \\ = 11! \\ = 3,99,16,800 \\\end{aligned} \end{equation} $$
(B) Only $6$ out of $8$ are selected.
Total arrangements is given by, $$\begin{equation} \begin{aligned} = \frac{{{}^8{P_6}}}{{(6)}} \\ = \frac{{8!}}{{6(8 - 6)!}} \\ = 3,360{\text{ ways}} \\\end{aligned} \end{equation} $$
Question 13. 20 delegates are invited for a conference. They sit for a dinner in a round table in such a way that three particular delegates always sit together. Find the possible ways of arranging them. Find the number of arrangements possible if two delegates always sit on either side of the host.
Solution: When three particular delegates always sit together, they can be considered as one single unit.
Thus there are $18$ units.
Therefore arranging them in, $$\begin{equation} \begin{aligned} = (18 - 1)! \\ = 17!{\text{ ways}} \\\end{aligned} \end{equation} $$
The three of them can be arranged in $3!$ ways.
Thus total number of ways is, $$ = (17!) \times (3!){\text{ ways}}$$
In the next case, again the three are considered as a single unit.
Since two of them sit on either sides of the host, there are $2$! ways of arranging them.
Thus total number of ways is, $$ = (17!) \times (2!){\text{ ways}}$$
Question 14. Find the number of ways in which $4$ Indians, $3$ Russians, $3$ Americans and $2$ Frenchmen can be seated in a circle, if the people of same nationality always sit together.
Solution: Let us considered $4$ Indians to be one unit $A$, $3$ Americans as another unit $B$, $3$ Russians to be another unit $C$ and the $2$ Frenchmen as a unit $D$.
Thus there are $(4-1)!$ ways of arranging these units.
Now,
$4$ Indians can be arranged in $4!$ ways,
$3$ Russians in $3!$ ways,
$3$ Americans in $3!$ ways and
$2$ Frenchmen in $2!$ ways.
Thus the total number of ways,
$$\begin{equation} \begin{aligned} = (4 - 1)!3!3!2!4! \\ = 2 \times {(3!)^3}4! \\ = 10368\;ways \\\end{aligned} \end{equation} $$
