**Permutation** (**Factorial function)-** We denote the product of the positive integers from 1 to n by n! and we read it as “factorial

It is expressed as below-

Note-

**Binomial coefficient**

Example:

**Permutation**

We call the arrangement of a set of n objects in a given order- permutation.

Any arrangement of any of these objects is r-permutation.

The number of permutations of n objects taken r- P(n, r)

Formula for P(n, r) is-

**Permutation with repetitions**

The number of permutations of *n *objects of which are alike is, so that we write it as

**Sampling with or without replacement**

Suppose we chose the samples with repetition, fro example if we draw a ball from a urn then we put back that ball in the urn and again we pick a ball and we continue the process, so this is the case of sampling with repetition

Then the product rule tells us that the number of such samples is-

And if we pick a ball from the urn and we do not put it back to the urn, then this is the case of sampling without replacement.

So that in this case the number of samples are-

**Example-**

**If we choose** t**hree cards** **one after the other from a 52-card deck. Find the number m of ways**

**This can be done: (a) with replacement; (b) without replacement.**

Sol.

(a) Each card can be chosen in 52 ways. Thus m = 52(52)(52) = 140 608.

(b) Here there is no replacement. Thus the first card can be chosen in 52 ways, the second in 51 ways, and the third in 50 ways. So that-

P(52, 3) = 52(51)(50) = 132 600

**Example-**

**There are 4 black, 3 green and 5 red balls. In how ma**ny ways we arrange them in a row**?**

**Solution: **Total number of balls = 4 black + 3 green + 5 red = 12

The black balls are alike,

The green balls are, and the red balls are alike,

The number of ways in which we can arrange the balls in a row =

**Circular permutations**

A circular permutation of *n *objects is an arrangement of the objects around a circle.

If the *n *objects are to be arranged round a circle we take an objects and fix it in one position.

Now the remaining (*n *– 1) objects can be arranged to fill the (*n *– 1) positions the circle in (*n *– 1)! ways.

Hence the number of circular permutations of *n *different objects = (*n *– 1)**!**