Permutations - RD Sharma Solved Problems.

 A permutation is an arrangement of objects in a definite order. The members or elements of sets are arranged here in a sequence or linear order. For example, the permutation of set A={1,6} is 2, such as {1,6}, {6,1}. As you can see, there are no other ways to arrange the elements of set A.

In permutation, the elements should be arranged in a particular order whereas in combination the order of elements does not matter. Also, read: Permutation And Combination

When we look at the schedules of trains, buses and the flights we really wonder how they are scheduled according to the public’s convenience. Of course, the permutation is very much helpful to prepare the schedules on departure and arrival of these. Also, when we come across licence plates of vehicles which consists of few alphabets and digits. We can easily prepare these codes using permutations.

A permutation is defined as an arrangement in a definite order of a number of objects taken, some or all at a time. Counting permutations are merely counting the number of ways in which some or all objects at a time are rearranged. The convenient expression to denote permutation is defined as “ nPr ”.

The permutation formula is given by,

Pr = n!/(n-r)! ; 0 ≤ r ≤ n

Where the symbol “!” denotes the factorial which means that the product of all the integers is less than or equal to n but it should be greater than or equal to 1.

Permutation When all the Objects are Distinct

There are some theorems involved in finding the permutations when all the objects are distinct. They are :

Theorem 1: If the number of permutations of n different objects taken r at a time, it will satisfy the condition 0 < r ≤ n and the objects which do not repeat is n ( n – 1) ( n – 2)……( n – r + 1), then the notation to denote the permutation is given by “ n Pr”

Theorem 2: The number of permutations of different objects “n” taken r at a time, where repetition is allowed and is given by nr .

Permutation When all the Objects are not Distinct Objects

Theorem 3: To find the number of permutations of the objects ‘n’, and ‘p’s are of the objects of the same kind and rest is all different is given as n! / p!

Theorem 4: The number of permutations of n objects, where p1 are the objects of one kind, p2 are of the second kind, …, pk is of the kth kind and the rest, if any, are of a different kind, then the permutation is given by n! / ( p1!p2!…Pk!)