Quantitative Analysis

Study Notes and Chapters for Quantitative Analysis - Online Preparation for Bank Exams

Introduction

Quantitative Analysis requires a lot of practice and hardwork since its an analytical subject. Before becoming an expert one should understand the basics of Quantitative Analysis for Government Bank Job Exams for SBI, RBI, RRB and IBPS. This section gives you all details about the basics of Quantitative Analysis. Here you can get the Free Notes and PDF downloads for Quantitative Analysis to prepare for the Bank PO and Clerical Jobs in India. Our experts are available incase you need any kind of online help to understand any topic or any kind of doubt clearing

Team - Laqshya ..

What you will learn

Our Main goal is to make Quantitative Analysis easy for every student to understand. In this section you will learn :

  • Quantitative Analysis indepth Knowledge
  • Improve your skills in Quantitative Analysis
  • Mastery to understand Quantitative Analysis
  • Became an exeprt in limited time
⇐ Back

Quantitative Aptitude Free-Study Notes

19. PERMUTATIONS AND COMBINATIONS

A. FUNDAMENTAL PRINCIPLES OF COUNTING

1. Multiplication Rule: If certain thing may be done in ‘m’ different ways and when it has been done, a second thing can be done in ‘n ‘ different ways then total number of ways of doing both things simultaneously = m × n.

Eg. if one can go to school by 5 different buses and then come back by 4 different buses, then total number of ways of going to and coming back from school = 5 × 4 = 20.

2. Addition Rule : It there are two alternative jobs which can be done in ‘m’ ways and in ‘n’ ways respectively then either of two jobs can be done in (m + n) ways.

Eg. if one wants to go school by bus where there are 5 buses or to by auto where there are 4 autos, then total number of ways of going school = 5 + 4 = 9.

AND => Multiply

OR => Add

3. Factorial : The factorial n, written as n! , represents the product of all integers from 1 to n both inclusive. To make the notation meaningful, when n = o, we define o! = 1.

Thus, n! = n (n – 1) (n – 2) ….. …3.2.1

Example : Find 5!, 4! and 6!

Solution : 5! = 5 × 4 × 3 × 2 × 1 = 120; 4! = 4 × 3 × 2 × 1 = 24; 6! = 6 × 5 × 4 × 3 × 2 × 1 = 720.

B. Permutation : The ways of arranging or selecting smaller or equal number of persons or objects from a group of persons or collection of objects with due regard being paid to the order of arrangement or selection, are called permutations.

Example : A group of persons want themselves to be photographed. They approach the photographer and request him to take as many different photographs as possible with persons standing in different positions amongst themselves. The photographer wants to calculate how many films does he need to exhaust all possibilities? How can he calculate the number?

In the situations such as above, we can use permutations to find out the exact number of films.

Solution : Let us explain, how the idea of permutation will help the photographer. Suppose the group consists of Mr. Suresh, Mr. Ramesh and Mr. Mahesh. Then how many films does the photographer need? He has to arrange three persons amongst three places with due regard to order. Then the various possibilities are (Suresh, Mahesh, Ramesh), (Suresh, Ramesh, Mahesh), (Ramesh, Suresh, Mahesh), (Ramesh, Mahesh, Suresh), (Mahesh, Ramesh, Suresh) and (Mahesh, Suresh, Ramesh ). Thus there are six possibilities. Therefore he needs six films. Each one of these possibilities is called a permutation of three persons taken at a time.

The different arrangements of a given number of things by taking some or all at a time, are called permutations.

Ex. 1.All permutations (or arrangements) made with the letters a, b, c by taking two at a time are: (ab, ba, ac, bc, cb).

Ex. 2.All permutations made with the letters a,b,c, taking all at a time are: (abc, acb, bca, cab, cba).

Number of Permutations: Number of all permutations of n things, taken r at a time, given by:

n Pr = n(n-1)(n-2).....(n-r+1) = n!/(n-r)!
Examples: (i) 6p2 = (6x5) = 30. (ii) 7p3 = (7x6x5) = 210.

Cor. Number of all permutations of n things, taken all at a time = n!

An Important Result: If there are n objects of which p1 are alike of one kind; p2 are alike of another kind; p3 are alike of third kind and so on and pr are alike of rth kind, such that (p1+p2+... .pr) = n.
Then, number of permutations of these n objects is: n! / (p1!).p2!)......(pr!)

C. Combinations : Each of the different groups or selections which can be formed by taking some or all of a number of objects, is called a combination.

We have studied about permutations in the earlier section. There we have said that while arranging, we should pay due regard to order. There are situations in which order is not important. For example, consider selection of 5 clerks from 20 applicants. We will not be concerned about the order in which they are selected. In this situation, how to find the number of ways of selection? The idea of combination applies here.

Definition : The number of ways in which smaller or equal number of things are arranged or selected from a collection of things where the order of selection or arrangement is not important, are called combinations.

Ex. 1. Suppose we want to select two out of three boys A, B, C. Then, possible selections are AB, BC and CA. Note that AB and BA represent the same selection.
Ex. 2. All the combinations formed by a, b, c, taking two at a time are ab, bc, ca.
Ex. 3. The only combination that can be formed of three letters a, b, c taken all at a time is abc.
Ex. 4. Various groups of 2 out of four presons A, B, C, D are: AB, AC, AD, BC, BD, CD.
Ex. 5. Note that ab and ba are two different permutations but they represent the same combination.

Number of Combinations: The number of all combination of n things, taken r at a time is:

n Cr = n! / (r!)(n-r)! = n(n-1)(n-2).....to r factors / r!

Note that: nCr = 1 and nC0 = 1.

An Important Result: nCr = nC(n-r).

Example :

(i) 11C4 = (11x10x9x8) / (4x3x2x1) = 330.
(ii) 16C13 = 16C(16-13) = 16x15x14 / 3! = 16x15x14/3x2x1 = 560.

 


Back to top