Today we’ll discuss about Permutation and Combination. This topic is can fetch you marks easily but you need to know the right concepts and types of questions to practice.

## All about Permutation and Combination Permutation

Permutations are different numbers of arrangements. Making words of 4 letters without repetition or with repetition is an example of permutation. Now suppose we have to make a word of 4 letters by taking any 4 letters from 10 letters A, B, C, D, E, F, G, H, I

☞ For example, permutations of 4 distinct objects out of  20 distinct objects can be written as

## Combination

A combination is a grouping or selection of all or part of a number of things without reference to arrangement of the things selected.
For example selection of three members from a family of ten, number of cricket matches in a group having 10 teams, number of triangles formed by joining any three points out of 20 points in plane etc.
☞ The symbol
represents total number of selections taking r different objects out of n different objects.

Suppose there are 5 students and a company wants to select any 3 of them. So the total number of ways = selection of 3 out of 5 = rejection of 2 out of 5.

## Word Formation

Word formation is based on fundamental principle of multiplication.

☞ For example, total number of words which can be formed by using the letters of the word “ACHILLES” will be 8!/2!, as the letter L appeared twice.

## Gap Method

Gap method is used if we have to arrange ‘n’ different things in such a way such that no two of the ‘r’ things are together. In this method we first place (n – r) things in a line so that there are (n – r + 1) gaps in between and at extreme ends. Now place ‘r’ things on these gaps so that no two of these things are together provided that r ≤ n – r + 1. Suppose T1,T2,T3,………… T10 are 10 distinct objects and we want that no two of T1,T2,T3 and T4 are together in the arrangement. First place remaining things T5,T6……… T10
_T5_T6_T7_T8_T9_T10_
So there are 7 gaps
Total number of ways = (7.6.5.4).6!

## Rank of a Word in Dictionary

If all the words which can be made by using n different letters are arranged in a dictionary in alphabetical order then position at which a particular word appears, is known as its rank. For example 24 words can be made by using letters of the word “RANK” without repetition. Now to calculate position of the word “RANK” we arrange the words in alphabetical order.

Hence position of the word RANK is:
3! + 3! + 3! + 1 + 1 = 20

## Sum of all the numbers which can be formed by using the digits without repetition

Suppose the digits are 1, 2, 3, 4. We know that total number of numbers which can be formed without repetition is 4! = 24. Now
Numbers with unit digit 1 will be 3! = 6
Numbers with unit digit 2 will be 3! = 6
Numbers with unit digit 3 will be 3! = 6
Numbers with unit digit 4 will be 3! = 6
Hence sum of the unit digits
= 6 × (1 + 2 + 3 + 4) = 60,
Similarly sum of the digits at tens, hundred’s and thousand’s place = 60
Sum of all the numbers
= 60 × 1 + 60 × 10 + 60 × 100 + 60 × 1000 = 66660
Sum of all the numbers which can be formed by using the ‘n’ non zero digits without repetition is (n – 1)! (sum of the digits) (111……n times).

## Circular Permutation

Number of circular permutations of n distinct objects = (n – 1)!
Number of circular permutations of n distinct objects =
if there is no difference between clock wise and anticlockwise permutations.

Practice Permutation and Combinations Questions