Question

What is circular permutation? give some examples also

Answer 1

See below:
Explanation:
A permutation is a method to calculate the number of ways we can take some number of objects and arranging them in some order. For instance, we can say that we have $5$ books and we want to put them on a shelf. The number of ways we can arrange those $5$ books is $5!=120$ ways.
We can also take a smaller grouping and arrange those. For instance, from the $5$ books, we want to take a group of $3$ of them and arrange them on a shelf. How many ways can we do that? That calculation is:
$5\times 4\times 3=60$ different ways. The permutation formula allows us to find that number directly:
$P_{n,k}=\frac{n!}{(n-k)!};n=$ population, k = picks
Note that we can express the two situations above as:
$P_{5,5}=\frac{5!}{0!}=5!=120$
and
$P_{5,5}=\frac{5!}{2!}=\frac{5\times 4\times 3\times 2!}{2!}=5\times 4\times 3=60$
So what is circular permutation?
Let's say that instead of books, we have $5$ people and they are going to sit at a table in a restaurant. What's different here is that instead of being in a line, the people are in a circle, and so having person A sitting in seat $1$ with person $B$ in seat $2$ on the right and person $E$ on the other side is the same as that same arrangement with person $A$ in seat $2$ (with $B$ in seat $3$ and $E$ in seat $1$). And seat $3$, and seat $4$, and seat $5$.
And so we end up with duplicates of the arrangements. On a circular table, $ABCDE$ is the same as $EABCD$ is the same as $DEABC$, and so on.
And so to eliminate the duplicates, we divide by the number of places. In this case, we get:
$\frac{5!}{5}=\frac{120}{5}=24$
Notice that with $5!=5\times 4\times 3\times \times 2\times 1$, we can also express this as:
$\frac{5!}{5}=4!=24$
Let's go one step further. We have $5$ people who walk into a restaurant but only $3$ are going to eat - two people will sit at the bar. In how many different ways can we arrange the $3$ people at the table?
We know that in our permutation, we have $\frac{5!}{2!}=3!=60$, so $60$ ways to arrange the $5$ people in groups of $3$ at the table. We then need to work with the table arrangement and recognize that $ABC$ is the same as $CAB$, so we divide by $3-$ the number of seats at the table, to get $20$.
We can list those out:
$ABC,ACB$
$ABD,ADB$
$ABE,AEB$
$ACD,ADC$
$ACE,AEC$
$ADE,AED$
$BCD,BDC$
$BCE,BEC$
$BDE,BED$
$CDE,CED$