Learn at My Place/Competitive Exams/CUET UG/Quantitative Aptitude/Modern Maths
CUET UG / Quantitative Aptitude / Modern Maths

Modern Maths for CUET UG: Permutation, Combination, Probability & Set Theory

Learn the high-yield modern maths block through concept-first notes, solved examples, and a timed practice route built around counting, probability, overlap logic, and survey-style questions.

10 Core Concept BlocksSolved Examples4 Sectional Sessions40-Question Mock
Go to NotesSolved ExamplesOpen Practice Tests
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Overview

Why Modern Maths Matters in CUET UG

Modern Maths is one of the cleanest scoring areas in CUET UG because it rewards pattern recognition more than long calculation. Once you identify whether the question is about arrangement, selection, overlap, or chance, the path becomes much shorter.

Permutation and combination build the counting logic, probability converts counting into likelihood, and set theory organizes overlap through Venn-style thinking. That is why these topics work best when studied together rather than as isolated formulas.

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Section A

Notes & Concept Builder

Identify the counting model first
1. Factorial and Counting Principles2. Permutation3. Combination4. Repeated Objects and Circular Arrangement5. Conditional Selection6. Probability Basics7. Complement, Addition, and Independence8. With and Without Replacement9. Set Theory10. Subsets and Proper Subsets

1. Factorial and Counting Principles

Modern Maths begins with systematic counting. Factorial compresses descending products and appears throughout permutation and combination.

n!=n×(n−1)×(n−2)×⋯×10!=1n!=n\times(n-1)\times(n-2)\times \dots \times 1 \qquad 0!=1

The multiplication principle is used when a task happens in stages. The addition principle is used when one among non-overlapping alternatives is chosen.

10-second shortcut: If the problem says “and then”, multiply. If it says “either this or that”, add.

2. Permutation

Permutation counts arrangements, so order matters. Seat order, password order, rank order, and medal order all fall into this bucket.

nPr=n!(n−r)!^nP_r=\frac{n!}{(n-r)!}

If 3 positions are to be filled from 7 candidates, think of 7 choices for the first position, then 6, then 5.

3. Combination

Combination counts selections, so order does not matter. Team, committee, panel, shortlist, and pair questions usually use combination.

nCr=n!r!(n−r)!^nC_r=\frac{n!}{r!(n-r)!}
10-second shortcut: Use symmetry:
nCr=nCn−r^nC_r=^nC_{n-r}
. That can make large-looking numbers very small.

4. Repeated Objects and Circular Arrangement

If identical objects are repeated, swaps among identical items do not create new arrangements. That is why we divide by repeated factorials.

Arrangements with repetition=n!p!q!r!…\text{Arrangements with repetition} = \frac{n!}{p!q!r!\dots}

For circular arrangement of distinct objects, rotations are treated as the same arrangement.

Circular arrangement of n distinct objects=(n−1)!\text{Circular arrangement of } n \text{ distinct objects}=(n-1)!

5. Conditional Selection

Many CUET questions add conditions such as exactly two girls, at least one boy, or a fixed member must be included. These are best handled through cases or complement.

For “at least one”, the complement is usually faster: total groups minus the groups that violate the condition.

6. Probability Basics

Probability converts counting into likelihood. Start with the ratio of favourable outcomes to total outcomes.

P(E)=favourable outcomestotal outcomesP(E)=\frac{\text{favourable outcomes}}{\text{total outcomes}}

Dice, coins, and cards all follow the same logic, but the sample space must be counted carefully.

7. Complement, Addition, and Independence

The complement rule often saves time in “at least one” questions.

P(E′)=1−P(E)P(E')=1-P(E)

If events are mutually exclusive, their overlap is zero. If they are independent, the probability of both is the product of their probabilities.

P(A∪B)=P(A)+P(B)−P(A∩B)P(A\cup B)=P(A)+P(B)-P(A\cap B)
P(A∩B)=P(A)P(B) for independent eventsP(A\cap B)=P(A)P(B) \text{ for independent events}

8. With and Without Replacement

Replacement keeps the denominator unchanged. Without replacement changes the denominator on the second draw, so the events become dependent.

Exam trap: Many wrong answers come from treating without-replacement draws as independent.

9. Set Theory

Set theory is the visual language of overlap. Union means combined coverage, intersection means common part, and complement means outside the set but inside the universal set.

n(A∪B)=n(A)+n(B)−n(A∩B)n(A\cup B)=n(A)+n(B)-n(A\cap B)

In survey questions, use the union formula first, then subtract from the total if the question asks for “neither”.

10. Subsets and Proper Subsets

Every element either belongs to a subset or does not. That is why a set with nn elements has 2n2^n subsets.

Number of subsets=2nNumber of proper subsets=2n−1\text{Number of subsets}=2^n \qquad \text{Number of proper subsets}=2^n-1
Solved Practice

Solved Examples

Try first, then open the reasoning
Exam Trap: Many wrong answers come from using permutation where combination is needed, or from forgetting overlap in probability and set questions.
Example 1. Evaluate $\frac{8!}{6!}$.
Cancel 6!6! to get 8×7=<strong>56</strong>8\times7=<strong>56</strong>.
Example 2. A student has 4 shirts and 3 caps. How many outfit pairs are possible?
Use the multiplication principle: 4×3=<strong>12</strong>4\times3=<strong>12</strong>.
Example 3. Find $^7P_3$.
Use permutation: 7×6×5=<strong>210</strong>7\times6\times5=<strong>210</strong>.
Example 4. How many ways can 4 friends sit in a row?
This is a straight-line arrangement of 4 distinct people, so 4!=244!=24.
Example 5. How many ways can 5 students sit around a round table?
For circular arrangement, use (5−1)!=<strong>24</strong>(5-1)! = <strong>24</strong>.
Example 6. How many arrangements of LEVEL are possible?
LEVEL has 5 letters with L repeated twice and E repeated twice. Count =5!2!2!=<strong>30</strong>=\frac{5!}{2!2!}=<strong>30</strong>.
Example 7. How many committees of 3 can be formed from 8 students?
Use combination: 8C3=<strong>56</strong>^8C_3=<strong>56</strong>.
Example 8. From 6 players, how many ways can captain and vice-captain be chosen?
Order matters, so use permutation: 6P2=<strong>30</strong>^6P_2=<strong>30</strong>.
Example 9. Find $^9C_7$ quickly.
Use symmetry: 9C7=9C2=9×82=<strong>36</strong>^9C_7=^9C_2=\frac{9\times8}{2}=<strong>36</strong>.
Example 10. From 5 boys and 4 girls, how many groups of 3 contain at least one girl?
Total groups =9C3=84=^9C_3=84. All-boy groups =5C3=10=^5C_3=10. Required =84−10=<strong>74</strong>=84-10=<strong>74</strong>.
Example 11. A fair die is rolled. What is the probability of getting an even number?
Even outcomes are 2, 4, and 6. So probability =36=<strong>12</strong>=\frac{3}{6}=<strong>\frac12</strong>.
Example 12. Two fair coins are tossed. Find probability of exactly one head.
Favourable outcomes HT and TH give probability =24=<strong>12</strong>=\frac{2}{4}=<strong>\frac12</strong>.
Example 13. Two dice are rolled. What is the probability that the sum is 8?
The favourable pairs are (2,6),(3,5),(4,4),(5,3),(6,2)(2,6),(3,5),(4,4),(5,3),(6,2), so probability =<strong>536</strong>=<strong>\frac{5}{36}</strong>.
Example 14. A card is drawn from a standard deck. What is the probability that it is not a heart?
Probability of a heart is 14\frac14, so not a heart is 1−14=<strong>34</strong>1-\frac14=<strong>\frac34</strong>.
Example 15. A box has 4 green and 6 yellow balls. Two draws are made with replacement. Probability both are green?
Probability =25×25=<strong>425</strong>=\frac25\times\frac25=<strong>\frac{4}{25}</strong>.
Example 16. If $A=\{2,4,6,8\}$ and $B=\{4,8,10\}$, find $A\cap B$.
The common elements are {4,8}\{4,8\}.
Example 17. If $A=\{1,2,3\}$ and $B=\{3,4,5\}$, find $A\cup B$.
Union contains all distinct elements, so {1,2,3,4,5}\{1,2,3,4,5\}.
Example 18. In a class, 24 students study English, 18 study Maths, and 7 study both. How many study at least one?
Use the union formula: 24+18−7=<strong>35</strong>24+18-7=<strong>35</strong>.
Example 19. How many subsets does a set with 4 elements have?
Number of subsets =24=<strong>16</strong>=2^4=<strong>16</strong>.
Example 20. How many proper subsets does a 5-element set have?
Proper subsets =25−1=<strong>31</strong>=2^5-1=<strong>31</strong>.
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Section B

The Test Zone

This chapter has a separate practice route with 4 sectional sessions of 10 questions each and a mixed 40-question mock. Every question runs on a 60-second timer so we train both concept clarity and exam speed.

Sectional Tests

4 focused sessions on counting, committee logic, probability, and set theory.

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Full Mixed Mock

40 questions blending permutation, combination, probability, and set theory in CUET style.

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Move straight from chapter-wise questions into a subject test, then loop back into weaker areas instead of ending the session here.

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