# Formula Sheets for General Aptitude (Part A): Permutation and combination

By Astha Singh|Updated : May 10th, 2023

General Aptitude Formula Sheets: During the preparation, the candidates study different formulas to solve problems, but at the last moment, these formulas might not be remembered by the candidates due to exam fear or pressure. We at BYJU'S Exam Prep do not want our students to lag anywhere during the preparation, so we have come up with a concept of a Formula Sheet that will help them revise the important formulas at the last moment. This formula sheet will be a short revision tool and contain only important formulas that need to be studied at the last minute to boost the score. Our experienced subject-matter experts have meticulously designed this CSIR NET General Aptitude Formula Sheet to provide you with the best authentic material.

In this article, we will cover the CSIR NET General Aptitude Most Important Formulas of Permutation and combination. Aspiring candidates can check all the most important formulas of the Mensuration for the last-minute revision. Scroll down the full article to find out!

## Formula Sheet On Permutation and combination

Fundamental Principle of Counting:
1. Rule of Sum:- A task is performed in m ways and another task is performed in n ways and both tasks cannot be performed simultaneously. So, either task can be accomplished in (m + n) ways.
2. Rule of Multiplication:- There are two tasks, A and B can be performed in m and n ways respectively. So, the number of different ways of doing both tasks A and B simultaneously is (m × n) ways.

Factorial: Factorial is a notation for multiplication of consecutive integers. The factorial is represented by the symbol ‘!’
n! = 1 × 2 × 3 × 4 × …. × (n – 1) × n (n! mean multiplication of first n natural numbers)
Ex: 4! = 1 x 2 x 3 x 4
n! = n × (n – 1)!
Ex: 4! = 4 x (4 – 1)! = 4 x 3!
Value of the few frequently used factorial.

0! = 1
1! = 1
2! = 2 x 1 = 2
3! = 3 x 2 x 1 = 6
4! = 4 x 3 x 2 x 1 = 24
5! = 5 x 4 x 3 x 2 x 1 = 120
6! = 6 x 5 x 4 x 3 x 2 x 1 = 720
7! = 7 x 6 x 5 x 4 x 3 x 2 x 1 = 5040 and so on.

Permutation: Permutation means arranging i.e. “selecting and ordering” one or more objects from the given certain objects (maybe alike or different). The number of permutations of n different objects taken r at a time is represented as : n(n – 1)(n – 2) ……… (n – r + 1) (where, 0 ≤ r ≤ n)n(n – 1)(n – 2) ……… (n – r + 1) (where, 0 ≤ r ≤ n)In Permutation and Combination, Combination is selection and Permutation is selection as well as arrangements.

Combination: Combination means selecting one or more objects from the given certain objects (maybe alike or different). The combination of n distinct objects taken r at a time is represented and calculated as: Here, r can be any positive integer less than or equal to n.
Ex: In a bag, there are 3 red, 4 black and 6 green balls. In how many different ways, we can select 2 red balls?
Sol: Required number of ways = 3C2
Ex: In a bag, there are 3 red, 4 black and 6 green balls. In how many different ways, we can select 2 balls such that none of them is red?
Sol: Required number of ways = (4 + 6)C2 = 10C2
Ex: In a bag, there are 3 red, 4 black and 6 green balls. In how many different ways, we can select 2 balls such that both are of the same colour?
Sol: Required number of ways = 3C2 + 4C2 + 6C2
Ex: In a bag, there are 3 red, 4 black and 6 green balls. In how many different ways, we can select 2 balls such that both are of different colours?
Sol: To select 2 balls of different colours, we can select in the combination of (red, black), (black, green), (green, red)
Required number of ways = (3 × 4) + (4 × 6) + (6 × 3) = 54 Ex: In how many different ways can the letters of the word 'ROSTED' be arranged?

Sol: Required number of ways = 6!

Permutation of Alike Objects: The number of permutations of n objects taken all at a time in which, p are alike objects of one kind, q are alike objects of second kind & r are alike objects of a third kind and the rest Ex: In how many different ways can the letters of the word 'MOTION' be arranged?

Sol: Here, the total number of letters = 6 O – 2 (alike) and M, T, I, N Ex: In how many different ways can the letters of the word 'FRUSTRATION' be arranged?
Sol: Here, the total number of letters = 11
R – 2, T – 2 and F, U, S, A, I, O, N Ex: In How many different words can be formed with the letters of the word 'POLICE' beginning with P?
Sol: In this case, we will arrange the other 5 digits (except P, because P is fixed) Hence, the required number of ways = 5!
Ex: In How many different words can be formed with the letters of the word 'POLICE' beginning with P and ending with E?
Sol: In this case, we will arrange the other 4 digits (except P and E as both are fixed) Hence, the required number of ways = 4!

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