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

Permutations, Combinations and Counting short notes for CAT: 57 study cards covering concepts, formulas, shortcuts and exam traps, plus solved practice 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.

The 4-Step Counting Method

A reliable CAT method

Define one outcome.
Example: one complete sandwich order.

Split into stages.
Type → bread → size → sauce choice.

Count choices at each stage.
Keep restrictions attached to the correct stage.

Multiply inside, add outside.
One route uses multiplication. Alternative routes use addition.

Memory hook: Outcome → Stages → Choices → Combine.

PYQ Pattern 1: Sandwich Orders with Optional Sauces

CAT PYQ Backbone

Sandwich order count

A cafeteria offers 5 sandwich types. For each type, choose 1 of 4 breads and small or large size. Optionally, add up to 2 out of 6 sauces.

Fixed stages

5 \times 4 \times 2 = 40

Sauce choices

^{6} C_{0} +^{6} C_{1} +^{6} C_{2}

Sauce count:

^{6} C_{0} +^{6} C_{1} +^{6} C_{2} = 1 + 6 + 15 = 22

Total:

5 \times 4 \times 2 \times 22 = 880

Answer:

880

different sandwich orders.

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 short notes in this unit

chapter

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

Permutations, Combinations and Counting short notes for CAT: 57 study cards covering concepts, formulas, shortcuts and exam traps, plus solved practice questi

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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