# Homework Help: What is the difference between the Permutations and combinations

1. Apr 22, 2010

### r-soy

Hi

I want simple explanation of the Permutations and combinations and wich one has condition and I want simple example to undersand it

I want your help

2. Apr 22, 2010

### statdad

At the simplest: permutations are used to count arrangements (order is important) and combinations are used when order does not matter.

In neither case (again, in simplest examples) is it allowed to repeat items.

Example: Consider the four letters A, B, C, D

Question 1: How many different arrangements of three letters are there (repetitions not allowed)

By listing we get:
ABC, ACB, ABD, ADB, ACD,ADC
BAC, BCA, BAD, BDA, BCD, BDC
CAB, CBA, CBD, CDB, CAD, CDA
DAB, DBA, DAC, DCA, DBC, DCB

If you count there are 24 of these (they are considered different because of the different orderings). The long phrase is this:
"There are 24 permutations of four letters selected three at a time"

While there is no universal standard for the mathematical notation, the ones I've listed below are the most common.

$$P(4,3) = 24, \quad P^4_3 = 24, {}^4P_3 = 24$$

The general formula for calculation can be written as

$$P(m,n) = \frac{m!}{(m-n)!} = m \cdot (m-1) \cdot \dots \cdot (m - n + 1)$$

where $m$ is the number of items from which you choose, and $n$ is
the number of items selected.

For combinations, order is not important, only the set of objects selected. Again, if you look at the letters A, B, C, D, the number of ways to select three at a time (no repetititions) is
four - the selections are
ABC, ABD, ACD, BCD

In notation

$$C(4,3) = 4, \quad {}^4C_3 = 4, C^4_3 = 4$$

and the general formula is

$$C(m,n) = \frac{m!}{n!(m-n)!} = \frac{P(m,n)}{n!}$$

Hope this helps.

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