Suppose there are 3 balls: Red, Green and Blue. In shortcut we can identify them as R,G and B. Now, how many was we can arrange these balls? Definitely in six ways.
RGB, RBG, GRB,GBR, BRG, BGR
Six ways. It can be found by 3! which is 3 x 2 x 1= 6.
But if we are told to arrange these 3 balls in such a way that we will pick only two at a time, then how many ways we can arrange or sort it. Let's see,
RG, RB, GB, GR, BG, BR
This is also in six ways and this arrangement or sorting is called Permutations.
Let's think about this 3 balls again. How many ways we can pick 2 balls by ignoring their orders.
RG, RB, BG
or
GR, BR, GB
Only 3 ways. Both time we ignored their internal ordering or arrangement because RG and GR is same, only difference is order. So, the number of ways we picked 2 balls from 3 balls is called Combination.
Permutations:
The ordered arrangements of a set of distinct objects is called Permutations. This ordered arrangement could be of some elements. So, an ordered arrangement of r elements from n distinct object is called r-permutation. It is denoted as nPr, nPr or P(n,r).
The number of r permutation of a set with n distinct element is
Because we know that n! = n.(n-1).(n-2).....(n-r+1). (n-r)!
Example:
How many different ways there are to select 3 winners(1st,2nd,3rd) from 100 different people?
Selecting 3 winners(1st,2nd,3rd) from 100 different people means selecting 3 ordered elements from a set of 100 different elements. Here 3 permutation of a set of 100 elements is P(100,3) = 100 x 99 x 98 = 970200
Combinations:
Unordered selection of some element from n distinct object is called Combinations. So, selecting r element from n distinct object by ignoring their order(no repetition) is called r combination. It is denoted as nCr, nCr or C(n,r).
The number of r combinations of a set with n elements, where n is a non-negative integer and r is an integer with 0 <= r <= n, equals
The number of r permutation of a set with n distinct element is
P(n,r)= n.(n-1).(n-2).....(n-r+1)
The first element of the permutation can be chosen in n ways because there are n elements at first. After that there will be (n-1) element will be left so, the second permutation can be done n-1 ways. Similarly, the third will be n-2 ways and so on. The permutation of rth element can be done in (n-r+1) ways.
P(n,r) = n.(n-1).(n-2).....(n-r+1)
Because we know that n! = n.(n-1).(n-2).....(n-r+1). (n-r)!
Example:
How many different ways there are to select 3 winners(1st,2nd,3rd) from 100 different people?
Selecting 3 winners(1st,2nd,3rd) from 100 different people means selecting 3 ordered elements from a set of 100 different elements. Here 3 permutation of a set of 100 elements is P(100,3) = 100 x 99 x 98 = 970200
Combinations:
Unordered selection of some element from n distinct object is called Combinations. So, selecting r element from n distinct object by ignoring their order(no repetition) is called r combination. It is denoted as nCr, nCr or C(n,r).
The number of r combinations of a set with n elements, where n is a non-negative integer and r is an integer with 0 <= r <= n, equals
The formula can also be re-written as follows:
Example:
How many ways are there to select 5 players from a 10 member tennis team?
The answer can given by the number of 5 combinations of a set with 10 elements. So the combination is:
C(10,5) = 10! / 5!5! = 252
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