# Permutation and Combination – Fundamental Principle of Counting

Tuesday, May 26th, 2020 For this series of articles, I am assuming CAT 2021 would be the first time you would be attempting CAT, which essentially implies that you are not well versed with the basic ideas behind the Quantitative Aptitude portion. In the latter half of the year, I would move to slightly more advanced topics as by then, you would also have moved to the advanced stage of preparation.

I have often seen students struggle with the topic – ‘Permutation & Combination’. As a matter of fact, I have even seen some faculties shy away from conducting those classes. This fear, for the lack of a better word, stems from the fact that the options are often very confusing. Even if you make a mistake, miss out a case, take a wrong factorial – the answer you obtain is invariably in the options. One of the most basic tips that I would like to give you is this – if you are not able to solve the question, look at the answer and then try to figure out the logic behind the answer.

I think one of the main reasons behind lot of students making mistakes in questions based on ‘Permutation & Combination’ is the fact that they start the chapter with the two formulas which are given below: It is wrong to start off with these formulas. It is very important to understand the logic behind these formulas. And that is precisely what I wish to achieve with the help of this post by talking about ‘Fundamental Principles of Counting’.

Rule of Product: If there are ‘m’ ways to do something and there are ‘n’ ways to do another, then the total number of ways of doing both things is ‘m x n’.

To elaborate this with an example, assume that you have 4 T-shirts and 2 Jeans. The total number of ways in which you can decide what to wear is 4 x 2 = 8.

In case you are wondering ‘Why is it 8?’, the logic is pretty simple. With every T-shirt, you have a choice between the two Jeans. This is illustrated below:

Choices of dress: T1J1, T1J2, T2J1, T2J2, T3J1, T3J2, T4J1 and T4J2

An assumption here is that you are not bothered with trivialities such as dressing-sense. Because if you are, then the decision of which jeans to wear with respect to a t-shirt will not be an independent decision. The formula of ‘m x n’ ways is valid if and only if the decisions are independent of each other.

In case the decisions are not independent, then you would have to take care of the restrictions which are applicable.

You can also see Tips and Tricks to Solve Para-Jumble Questions for CAT Exam

Rule of Sum: If there are ‘m’ ways to do something and there are ‘n’ ways to do another and we cannot do both at the same time, then there are ‘m +n’ ways to choose one of the actions.

To elaborate this with an example, assume that you have 5 Formal Shoes and 3 Cowboy Boots. The total number of ways in which you can decide your footwear is 5 + 3 = 8.

In case you are wondering ‘Why is it 8?’, the logic is pretty simple. You can either wear Formal Shoes or Cowboy Boots but not both. The choices are illustrated below.

Choices of footwear: FS1, FS2, FS3, FS4, FS5, CB1, CB2 and CB3

A slightly more complicated example on the same would go something like this.

Question 1: You have 4 T-shirts, 2 Jeans, 6 Sarees, 5 Formal Shoes and 3 Cowboy boots. In how many ways can you decide what to wear?

–          The answer for this is (4 x 2 + 6) x (5 + 3) = 14 x 8 = 112 ways.

I hope the logic behind the answer would be clear to you by now.

Continuing with the same idea, try to answer this question.

Question 2: You have 50 students in a class and you have to select three out those for the posts of President, Vice-President and General Secretary. In how many ways can you do that?

–          The President can be any one of the 50 students. Suppose you choose X.

The Vice-President can be any of the remaining 49 students (Not X). Suppose you choose Y.

The General Secretary can be any of the remaining 48 students (Not X or Y).

So, the total number of ways in which you can decide the students for the positions are = 50 x 49 x 48.

Question 3: In how many ways can you select and arrange ‘r’ items out of ‘n’ distinct items?

The 1st item can be selected in ‘n’ ways.

The 2nd item can be selected in ‘n – 1’ ways.

The 3rd item can be selected in ‘n – 2’ ways.

The rth item can be selected in ‘n – r + 1’ ways.

So, the total number of ways of selecting and arranging ‘r’ items out of ‘n’ distinct items is: As you can realize, this is a difficult formula to remember. To take care of the same, multiply (n-r)! to both the numerator and the denominator.  Does the above formula look familiar? If not, just scroll up and see what n P r is.

Question 4: In how many ways can you arrange ‘r’ items?

–          The 1st item can be selected in ‘r’ ways.

The 2nd item can be selected in ‘r – 1’ ways.

The 3rd item can be selected in ‘r – 2’ ways.

The rth item can be selected in ‘r – (r – 1)’ ways or simply put, 1 way.

So, the total ways of arranging ‘r’ items is: Question 5: In how many ways can you select ‘r’ items out of ‘n’ distinct items?

–          From Question 3, I know that the number of ways of selecting and arranging is n P r.

From Question 4, I know that the number of ways of just arranging is r!

Selecting and Arranging are independent decisions, so The above equation not only gives us the formula for n C r, but it also gives us a very important relationship n P r = n C r x r!

I hope with the help of this post, the logic behind n P r and n C r would have become clear to you and you would not make a mistake in the same area ever again.

## Other posts related to Quantitative Aptitude – Modern Maths

### CAT Questions related to Quantitative Aptitude – Modern Maths

All questions from CAT Exam Quantitative Aptitude – Modern Maths
Quantitative Aptitude – Modern Maths – Progressions – Q1: If a1 = 1/(2*5), a2 = 1/(5*8), a3 = 1/(8*11),……, then a1 + a2 +……..+ a100 is
Quantitative Aptitude – Modern Maths – Progressions – Q2: An infinite geometric progression a1, a2, a3,… has the property that an = 3(a(n+ l) + a(n+2) +….) for every n ≥ 1. If the sum a1 + a2 + a3 +……. = 32, then a5 is
Quantitative Aptitude – Modern Maths – Progressions – Q3: Let a1, a2, a3, a4, a5 be a sequence of five consecutive odd numbers. Consider a new sequence of five consecutive even numbers ending with 2a3.
Quantitative Aptitude – Modern Maths – Progressions – Q4: Let a1, a2,……..a3n be an arithmetic progression with a1 = 3 and a2 = 7. If a1 + a2 + ….+a3n = 1830, then what is the smallest positive integer m such that m (a1 + a2 + …. + an ) > 1830?
Quantitative Aptitude – Modern Maths – Progressions – Q5: If the square of the 7th term of an arithmetic progression with positive common difference equals the product of the 3rd and 17th terms, then the ratio of the first term to the common difference is
Quantitative Aptitude – Modern Maths – P&C – Q1: How many four digit numbers, which are divisible by 6, can be formed using the digits 0, 2, 3, 4, 6, such that no digit is used more than once and 0 does not occur in the left-most position?
Quantitative Aptitude – Modern Maths – P&C – Q2: In how many ways can 8 identical pens be distributed among Amal, Bimal, and Kamal so that Amal gets at least 1 pen, Bimal gets at least 2 pens, and Kamal gets at least 3 pens?
Quantitative Aptitude – Modern Maths – P&C – Q3: In how many ways can 7 identical erasers be distributed among 4 kids in such a way that each kid gets at least one eraser but nobody gets more than 3 erasers?
Quantitative Aptitude – Modern Maths – P&C – Q4: Let AB, CD, EF, GH, and JK be five diameters of a circle with center at O. In how many ways can three points be chosen out of A, B, C, D, E, F, G, H, J, K, and O so as to form a triangle?
Quantitative Aptitude – Modern Maths – Set Theory – Q.1 If among 200 students, 105 like pizza and 134 like burger, then the number of students who like only burger can possibly be?
Quantitative Aptitude – Modern Maths – Progressions – Q.2 Let x, y, z be three positive real numbers in a geometric progression such that x < y < z. If 5x, 16y, and 12z are in an arithmetic progression then the common ratio of the geometric progression is? Quantitative Aptitude – Modern Maths – Set Theory – Q.3 Each of 74 students in a class studies at least one of the three subjects H, E and P. Ten students study all three subjects, while twenty study H and E, but not P. Every student who studies P also studies H or E or both. If the number of students studying H equals that studying E, then the number of students studying H is?
Quantitative Aptitude – Modern Maths – P&C – Q.4 How many numbers with two or more digits can be formed with the digits 1,2,3,4,5,6,7,8,9, so that in every such number, each digit is used at most once and the digits appear in the ascending order?
Quantitative Aptitude – Modern Maths – P&C – Q.5 In a tournament, there are 43 junior level and 51 senior level participants. Each pair of juniors play one match. Each pair of seniors play one match. There is no junior versus senior match. The number of girl versus girl matches in junior level is 153, while the number of boy versus boy matches in senior level is 276. The number of matches a boy plays against a girl is?
Quantitative Aptitude – Modern Maths – Progressions – Q.6 The arithmetic mean of x, y and z is 80, and that of x, y, z, u and v is 75, where u=(x+y)/2 and v=(y+z)/2. If x ≥ z, then the minimum possible value of x is?
Quantitative Aptitude – Modern Maths – Set Theory – Q.7 For two sets A and B, let AΔB denote the set of elements which belong to A or B but not both. If P = {1,2,3,4}, Q = {2,3,5,6,}, R = {1,3,7,8,9}, S = {2,4,9,10}, then the number of elements in (PΔQ)Δ(RΔS) is?
Quantitative Aptitude – Modern Maths – Set Theory – Q.8 If A = {6^2n -35n -1: n = 1,2,3,…} and B = {35(n-1) : n = 1,2,3,…} then which of the following is true?
Quantitative Aptitude – Modern Maths – Sequence and Series – Q.9 Let t1, t2,… be real numbers such that t1+t2+…+tn = 2n2+9n+13, for every positive integer n ≥ 2. If tk=103, then k equals?
Quantitative Aptitude – Modern Maths – Progressions – Q.10 Let a1, a2, … , a52 be positive integers such that a1 < a2 < … < a52. Suppose, their arithmetic mean is one less than the arithmetic mean of a2, a3, …, a52. If a52 = 100, then the largest possible value of a1 is?