The number of circular permutations of n different things taken all at a time is (n – 1)!

“n” different things can be arranged in “n” different places in a circular shape. Let the required number of circular permutations be denoted by k.

Each circular permutation formed from n objects gives n different similar permutations.

Therefore, from the k circular permutations, the number of similar permutations formed is k∗n.

But n! similar permutations can be from n different objects.

So k∗n = n!

⇒k = n! / n = (n – 1)!.

Thus, the required number of circular permutations is (n – 1)!

**Examples**

**Solution:**

From the theorem, the number of permutations of sitting is (6 –1)! = 5! = 120

If clockwise and anticlockwise directions are not considered, the number of permutations is 120/2 = 60.

**Solution:**

The arrangement of pearls in a necklace is in either the clockwise direction or the anticlockwise direction. Therefore, we cannot consider the direction in this arrangement.

∴ The number of ways of arranging the pearls is (7 – 1)!/2 = 360.

The number of permutations of “n” dissimilar objects taken “r” at a time, when the repetition of objects is allowed any number of times is nr

**1. There are nine objects and nine boxes. Out of nine, five objects cannot fit into three small boxes. How many arrangements can be made, such that each object can be put in only one box?**

**Solution:**

Since five objects cannot fit into three small boxes,
these five objects can be put in the remaining six boxes
in ^{6}p5 ways

The remaining four objects can be put in ^{4}p4 ways.

∴ The total number of arrangements

= ^{6}p5 x ^{4}p4

= (6∗5∗4∗3∗2∗1)∗(4∗3∗2∗1)

= 720∗24

= 17280.

There are “n” objects, out of which m (< n) objects are identical and the remaining are dissimilar. Using all of the objects, the number of permutations from “n” objects is n! / m!

**1. Six identical coins are arranged in a row. Find the number of arrangements in which four are heads and two are tails.**

**Solution:**

To solve this problem, we can think of finding the number of permutations from six objects in which four are of one type (heads) and two are of another type (tails).

∴ The required number of ways

** ** 6! 6 **∗**5**∗**4**∗**3**∗**2**∗**1

** **——— = ————————— = 15

** **4!**∗**2! (4**∗**3**∗**2**∗**1)(2**∗**1)

**2. Find the number of numbers formed from 0, 1, 2, 3, 4, 5 in which a digit cannot be repeated. Out of these numbers, find how many numbers that we can get that are above 3000.**

**Solution:**

Note that no number starts with the digit “0”.

Writing six digit numbers excepting ”0” in the first place, the remaining five digits can be written in the first place in ^{5}p1 = 5 ways.

After filling the first place we are left with five digits, including “0”.

So the remaining five places can be filled with five digits in ^{5}p5 = 5! ways

∴ The number of numbers formed with six digits = 5∗5! = 600.

Similarly,

5 digit numbers = 5 *5p^{4} = 600

4 digit numbers = 5 *^{5}p3 = 300

3 digit numbers = 5 *^{5}p2 = 100

2 digit numbers = 5 *^{5}p1 = 25

1 digit numbers = 5 *^{5}p0 = 5

∴ The total number of non–zero numbers

= 600 + 600 + 300 + 100 + 25 + 5 = 1630.

Find the number of numbers greater than 3000.

We first find the total number of six-digit numbers and five-digit numbers and then add these to the number of numbers formed with four digits in which 3, 4, 5 are in the first place.

The number of numbers in which 3 or 4 or 5 are in the first place = 3*^{5}p3 = 180

∴ The number of numbers greater than 3000

= 600 + 600 + 180 = 1380

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