Circular Permutations
The arrangements we have considered so far were linear.
There are arrangements in closed loops also, called as circular arrangements.
Suppose n persons (a1, a2, a3,…,an)
are to be arranged around a circular table. There are n! ways in which they can
be arranged in a row. On the other hand, all the linear arrangements depicted
by .
a1, a2, a3, ……..., an.
an, a1, a2,……….,an
– 1.
an – 1, an, a1, a2…..an
– 2.
…………………….
a2, a3, a4,……….,a1.
will lead to the same arrangement for a circular table.
Hence each circular arrangement corresponds to n linear arrangements (i.e. in a
row). Hence the total number of circular arrangements of n persons is 
= (n –
1)!.
In other words, the arrangement (permutation) in a row
has a beginning and an end, but there is nothing like beginning or end in
circular permutation. Thus, in circular permutation, we consider one object as
fixed and the remaining objects can be arranged in (n – 1)! ways (as in the
case of arrangement in a row).
Distinction
between clockwise and anti-clockwise Arrangements
Consider the following circular arrangements: In figure
I, the order is clockwise whereas in figure II, the order is anti-clock wise.
These are two
different arrangements. When distinction is made between the clockwise and the
anti-clockwise arrangements of n different objects around a circle, then the
number of arrangements = (n – 1)!.
But if no distinction is made between the clockwise and
the anti-clockwise arrangements of n different objects around a circle, then
the number of arrangements is 1/2(n – 1)!
For an example, consider the arrangements of beads (all
different) on a necklace as shown in figures A and B.
Look at (A) having 3 beads x1,x2,x3
as shown. Flip (A) over on its right. We get (B) at once. However, (A) and (B)
are really the outcomes of one arrangement but are counted as two different
arrangements in our calculation. To nullify this redundancy, the actual number
of different arrangements is (n-1)!/2. 
Remaks:
When the positions are numbered, circular arrangement is
treated as a linear arrangement.
In a linear arrangement, it does not make difference
whether the positions are numbered or not.
Example.1
Consider 23 different coloured beads in a necklace. In
how many ways can the beads be placed in the necklace so that 3 specific beads
always remain together?.
Solution:
By
theory, let us consider 3 beads as one. Hence we have, in effect, 21 beads,
'n' = 21. The number of arrangements = (n-1)! = 20!.
Also, the number of
ways in which 3 beads can be arranged between themselves is 3! = 3 x 2 x 1 = 6.
Thus the total
number of arrangements = (1/2). 20!. 3!.
Example.2
In how many ways 10 boys and 5 girls can sit
around a circular table so that no two girls sit together.
Solution
10 boys can be seated in a circle in 9! ways. There are
10 spaces inbetween the boys, which can be occupied by 5 girls in 10p5
ways. Hence total number of ways .
= 9! 10p5
= 
.
