Combination is for selection where as Permutation
is for selection and arrangement (putting in order). Order may be linear like
“right to left” or “left to right”, or may be circular like “clockwise” or
“anticlockwise”.

We consider one example of selecting 2 persons from
A,B,C.

We can select A&B or B&C or A&C – total

**3 possibilities**. There ends combinations.
If we need to arrange the selected persons, we will
get the following possibilities- AB,BA,BC,CB,AC,CA – total

arrangement. Got the difference between the two? That’s it!

**6 possibilities**. These are permutations. Here we have observed that permutations include selection andarrangement. Got the difference between the two? That’s it!

The above concept can be formulised as below:

No. Of way of selecting 2 out of 3 persons = 3C2
= (3!)/ (3-2)! (2!) = (3*2*1)/ (1)(2*1) =

**3**
No. Of ways of arranging 2 from given 3 persons = 3P2
= (3!)/ (3-2)! = (3*2*1)/ (1) =

**6****Number of possibilities of selecting r items out of n items = n**

**C**

**r**

**Number of possibilities of arranging r items out of n items = n**

**P**

**r**

**n**

**C**

**r = (n!)/ (n-r)! (r!)**

**n**

**P**

**r =(n!)/ (n-r)!**

Some useful deductions here:

Prior to proceeding for the deductions, I will
brief on factorial. Factorial is defined for all whole numbers. ie., n=0,1,2,3,....

n! = n * (n-1) * (n-2) *(n-3) *...3 * 2 * 1

n! = n * (n-1)!

0! = 1; 1! = 1;

nC0 = n!/(n-0)! 0! = 1; nC1 = n!/(n-1)! 1!
= n; nCn
= n!/(n-n)! n! = 1

nCr = nC(n-r)

nPr = nCr * (r!) = nCr
* rPr

nPr is generally greater than nCr excepting the case
where r=1, in which case nPr becomes equal to nCr

__“Repetitions are allowed”:__
Up to now we discussed about no-repetitions case. Now we will see the
case “if repetitions are allowed”. For that we will take the example of forming
different numbers using given digits:

Given numbers are 6, 7, 8. We need to form different 2-digit numbers
(here repetitions are allowed).

The possibilities are:

66, 77, 88 – repetition cases

67, 76, 68, 86,78, 87 – no
repetition cases

- Total
possibilities are “

**9**”. Here we are considering**permutations**(as we have taken care of the order of digits for example 78 and 87). But the earlier nPr formula won’t work here (observe 3P2 =6 but not 9) as repetitions are allowed here. The formula here is**n**. (Observe 3^{r}^{2}= 9)**Number of possibilities of arranging r items out of n items if repetitions are allowed = n**

^{r}

__Circular Permutations:__
Why not “circular combinations”? Its absurd because the word “circular”
here hints some arrangement. “Arrangement” indicates

**permutations**only but not it’s counter-part.
Find all possibilities of arranging 3 persons A,B,C around a circular
table?

In circular permutations, relative positions only are considered but not
exact positions. For that reason, all the three possibilities -ABC,BCA and CAB
comes under category-1 and similarly ACB,CBA and BAC comes under category-2. So we get only two possibilities as mentioned
in the diagram.

The formula here is (n-1)!

**The number of possibilities of arranging n-different things in circular order = (n-1)!**

If the things considered are identical, then clock-wise and
anti-clockwise won’t matter. And the number of possibilities gets halved.

**The number of possibilities of arranging n-identical things in circular order = (n-1)!/2**

## No comments:

## Post a Comment