Posted on - 20-02-2017

IIT JEE

The arrangements we have considered so far were linear. There are arrangements in closed loops also, called as circular arrangements.

Suppose n persons (a_{1}, a_{2}, a_{3},…,a_{n})
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 .

a_{1}, a_{2}, a_{3}, ……..., a_{n}.

a_{n}, a_{1}, a_{2},……….,a_{n
– 1}.

a_{n – 1}, a_{n}, a_{1}, a_{2}…..a_{n
– 2}.

…………………….

a_{2}, a_{3}, a_{4},……….,a_{1}.

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 x_{1},x_{2},x_{3}
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.

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.

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?.

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!.

In how many ways 10 boys and 5 girls can sit around a circular table so that no two girls sit together.

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 ^{10}p_{5}
ways. Hence total number of ways .

= 9! ^{10}p_{5}
= .