Permutation and combination

Permutation

Let us arrange three types of fruits namely  Apple, banana and mango in all possible ways(each is different).
   Then we will get  six different varieties.
 Apple, banana, mango;
 Apple, mango, banana;
 Banana, mango, apple;
 Banana, apple, mango;
 Mango, apple, banana;
 Mango, banana, apple;

All are different!

The arrangement of things like this are known as Permutation.

 Definition: 

Permutation is arrangement of things in all possible ways.

 In permutation the order of things is  considered.

     As for example , let, we have to form a number  consisting of three digits using the digits 1,2,3,4 . To form this number the digits have to be arranged  in some order. Different numbers will get formed depending upon the order in which we arrange the digits. In this way , each arrangement of the digits is a permutation.

  Again , let , there are three prizes and nine participants in a competition.

 We are to distribute the prizes among the top three(first, second and third) participants. Then we are to choose three people out of nine. Now, the first winner can be chosen in 9 different ways. The second winner can be chosen in 8 different ways. And  the third winner  can be chosen in 7 different ways. Thus we have total 9×8×7 different ways to choose three winners from a set of 9 participants.

   We know that, 9!=9×8×7×...×2×1.

Now , 9×8×7=9!/6!. That is, 9!/(9-3)!.

   In general, there are n!/(n-k)!  different ways to arrange k elements out of n elements in some order. Generally, it is denoted by P(n,k).

  So, P(n,k) = n!/(n-k)!.

    

Combination

 Let we are to select 11 players out of 15 players to form a cricket team. We can select any 11 of the 15 players randomly. Here if we change the order of the players the team does not changes. So, in combination order is not considered.

Definition:

Combination is the selection of things. 

In combination  the order of things is not considered.

As for example, let we are to distribute 3  prizes(same) to 3 winners(first, second and third) out of 9 participants. Since prize is same for all, the order(first, second, third) does not matter. Now, we can select 3 winners from 9 participants in P(9,3) different ways. But here order is considered. So, if we does not consider the order we have total P(9,3)/3! ways.

  Thus , we can select 3 winners from 9 participants in P(9,3)/3! ways. The order is not considered here. This is a good example of combination. 

  Generally, combination of k elements out of n elements is denoted by C(n,k).

 C(n,k)= n!/{(n-k)!×k!} = P(n,k)/k!.

 This is the basic concept of permutation and combination.

 Some interesting problems on permutation and combination:

Problem 1:

 How many words can be formed using any four letters from the word "SPRITE" ?

    

   Here, the word "SPRITE" contains six different letters. We are to form words choosing four letters out of these six letters in all possible ways. So, we are to choose four letters from six letters in all possible different ways. Therefore, the number of ways is equals to P(6,4) = 6!/(6-4)! =6!/2!=360.

Finally, 360 words can be formed by choosing four letters from six letters of the word "SPRITE".

 Problem 2:

  Suppose, there are 30 students in a class. We are to form  quiz  teams of 5 students from these  30 students. How many different teams can be formed?

Here , we are to form teams of 5 students from 30 students in all possible different ways. So, the possible number of teams are equals to C(30,5) = 30!/{5!×(30-5)!} = 142506.

  

Note:

(1) P(n,r)= P(n-1 , r) +{r× P(n-1, r-1)}.

(2) C(n,r)+C(n,r-1)=C(n+1,r).

(3) C(n,r) ÷ C(n,r-1) =(n-r+1)/r.

(4) C(n,r) =C(n, n-r).

(5) If C(n,p) = C(n,q) then, p+q=n (p not equals to q).

(6) C(n,1)+C(n,2)+C(n,3)+...+C(n,n)=2ⁿ -1.

   

If you find out any incorrect information or know anything more about this , please write it in the comment section!