Permutations, Combinations and Counting Notes for CAT: Concepts, Formulas, Worked Examples & Practice

Permutations, Combinations and Counting notes for CAT: 65 study cards covering concepts, formulas, shortcuts and exam traps, plus solved practice questions.

Chapter Roadmap: Permutations, Combinations and Counting

Modern Maths Journey

Permutations, Combinations and Counting

In this chapter, CAT usually checks whether you can convert a word problem into a clean count. The full chapter has 11 own-course PYQs out of 1002 total course questions.

🧩

t1. Counting Arrangements and Product Rule

Learn the base engine: multiply choices across stages, add mutually exclusive routes.

2 PYQs
support-core

🔢

t2. Digit Formation and Non-Repeating Numbers

Count numbers under digit restrictions, zero restrictions, and non-repetition.

6 PYQs
highest weight

🎈

t3. Distribution of Identical Objects

Distribute identical items under lower-bound and parity conditions.

2 PYQs
moderate

✅

t4. Selection with Restrictions

Handle must-include, must-exclude, and cannot-appear-together conditions.

1 PYQ
low-repeat

By the end: you should be able to look at a counting problem and immediately decide: multiply choices, add cases, or break the problem into stages.

Topic Hero: Counting Arrangements and Product Rule

🧮

Selected Topic

Counting Arrangements and Product Rule

The main skill is simple: do not list outcomes; instead, break the action into stages and count choices at each stage.

Stage 1
choices

Stage 2
choices

Stage 3
choices

CAT relevance

This topic has 2 direct PYQs. Its own frequency is moderate, but it is the base logic behind many counting questions.

The Product Rule: When Choices Happen One After Another

Product Rule

Use it when an outcome is built step-by-step.

If a task has stages with

a

b

, and

c

choices, then:

Total ways = a \times b \times c

Keyword feeling: choose this and then choose that.

Mini example: 5 sandwich types, 4 breads, 2 sizes gives

5 \times 4 \times 2 = 40

base orders.

Multiply for AND, Add for OR

The Counting Switch

AND → ×

One complete outcome needs all stages.

Bread and size

4 \times 2

OR → +

The final outcome can happen through separate cases.

Via Q or via R

case 1

+

case 2

Rule: multiply inside a route; add between non-overlapping routes.

Permutations, Combinations and Counting: Solved Questions with Step-by-Step Explanations (2 Problems)

Question 1 · Quantitative Ability NAT

Thirty identical candies are to be given to five children. Every child must receive some candies, and no child may receive an odd number of candies. How many distributions are possible?

Correct Answer:

1001

Step-by-Step Solution

Key idea: this is a parity-restricted identical-object distribution question. The phrase "some candies" means each child gets at least one, and "no child may receive an odd number" means each share is even. So each share must be a positive even number.

Step 1: Let child $i$ receive $x_{i}$ candies. We need

$x_{1} + x_{2} + x_{3} + x_{4} + x_{5} = 30$ ,

where each $x_{i}$ is even and positive.

Step 2: Use the substitution $x_{i} = 2 y_{i}$ . Since $x_{i}$ is positive even, $y_{i}$ must be a positive integer: $y_{i} \geq 1$ .

Step 3: Substitute into the total:

$2 y_{1} + 2 y_{2} + 2 y_{3} + 2 y_{4} + 2 y_{5} = 30$ .

Divide by $2$ :

$y_{1} + y_{2} + y_{3} + y_{4} + y_{5} = 15$ .

Step 4: Count positive integer solutions. For $5$ positive variables summing to $15$ , the number is

$(5 - 1 15 - 1) = (4 14)$ .

Step 5: Compute:

$(4 14) = \frac{14 \times 13 \times 12 \times 11}{4 \times 3 \times 2 \times 1} = 1001$ .

Answer: $1001$ .

Common trap: if you allow $y_{i} = 0$ , you are allowing a child to receive zero candies. That violates "every child must receive some candies".

Question 2 · Quantitative Ability NAT

Twelve identical stickers and seven identical badges are to be distributed among four children. Each child must receive at least $2$ stickers. Badges have no restriction: a child may receive zero badges. How many ways are there to distribute both types of items?

Correct Answer:

4200

Step-by-Step Solution

Key idea: this is a multiple-identical-item distribution question, recognisable because two different item types are being distributed to the same children. Since the sticker distribution does not restrict the badge distribution, count the two problems separately and multiply.

Step 1: Set up the sticker condition. Let $x_{i}$ be the number of stickers received by child $i$ . We need

$x_{1} + x_{2} + x_{3} + x_{4} = 12$ , with $x_{i} \geq 2$ .

Step 2: First satisfy the minimum. Give each child $2$ stickers. This uses $4 \times 2 = 8$ stickers, leaving $12 - 8 = 4$ stickers to distribute freely.

Step 3: Count the remaining sticker distributions. The number of non-negative solutions for $4$ identical stickers among $4$ children is

$(4 - 1 4 + 4 - 1) = (3 7) = 35$ .

Step 4: Count the badge distributions. Seven identical badges among four children with zero allowed gives

$(4 - 1 7 + 4 - 1) = (3 10) = 120$ .

Step 5: Multiply independent counts:

$35 \times 120 = 4200$ .

Answer: $4200$ .

Common trap: combining stickers and badges into one stars-and-bars count treats different item types as if they were identical. They are separate identical-object problems, so multiply the separate counts.

More notes in this unit

chapter

Permutations, Combinations and Counting Notes for CAT: Concepts, Formulas, Worked Examples & Practice

Permutations, Combinations and Counting notes for CAT: 65 study cards covering concepts, formulas, shortcuts and exam traps, plus solved practice questions.

A question from this chapter

Question 1

Thirty identical candies are to be given to five children. Every child must receive some candies, and no child may receive an odd number of candies. How many distributions are possible?

Question 2

Twelve identical stickers and seven identical badges are to be distributed among four children. Each child must receive at least 2 stickers. Badges have no restriction: a child may receive zero badges. How many ways are there to distribute both types of items?

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