6.4 Combinations – Class 11 Mathematics

Introduction

In many real-life and exam-based problems, the order of selection does not matter. For example:

• Choosing 3 students from a class of 10 for a project team.
• Selecting 5 cards from a deck of 52.
• Choosing a committee from a group of people.

In such cases, we deal with combinations rather than permutations.

• Permutation → Arrangement (order matters).
• Combination → Selection (order does not matter).

Definition of Combinations

The number of ways of selecting r objects from n distinct objects when the order does not matter is called a combination.

It is denoted by nCr (read as “n choose r”).

Formula for Combinations

where:

• n! = factorial of n
• r = number of objects chosen

Understanding the Difference (Permutation vs Combination)

Suppose we have letters A, B, C.

• If we arrange 2 letters at a time:
  • Permutations = AB, BA, AC, CA, BC, CB → 6
• If we just select 2 letters (order not important):
  • Combinations = AB, AC, BC → 3

So, permutations count arrangements, combinations count selections.

Relation between them:

Properties of Combinations

1. nC0 = 1 (choosing nothing = 1 way).
2. nCn = 1 (choosing all = 1 way).
3. nCr = nC(n-r) (symmetry property).
4. Sum property:

nCr+nC(r−1)=(n+1)CrnCr + nC(r-1) = (n+1)CrnCr+nC(r−1)=(n+1)Cr

(This is the basis of Pascal’s Triangle).

Examples (Step-by-Step)

Example 1: Simple Selection

How many ways can 3 students be chosen from 10?

Solution:
n = 10, r = 3

10C3 = 10! / (3! × 7!) = (10 × 9 × 8) / 6 = 120

Example 2: Committee Selection

From 8 men and 5 women, a committee of 3 people is to be formed. In how many ways can this be done?

Solution:
Total = 13 people
Choose 3 = 13C3 = (13 × 12 × 11) / 6 = 286

Example 3: Cards Problem

How many ways can 5 cards be chosen from a deck of 52?

Solution:
52C5 = 52! / (5! × 47!) = 2,598,960

Example 4: Using Symmetry

Evaluate 12C8 without expanding factorials.

Solution:
By property: nCr = nC(n-r)
12C8 = 12C4 = (12 × 11 × 10 × 9) / 24 = 495

Example 5: Password Example

How many ways can a 4-digit pin be chosen from digits 0–9 if repetition is not allowed and order does not matter?

Solution:
Choosing 4 digits out of 10 → 10C4 = 210

Advanced Examples (Competitive Exam Level)

Example 6: Mixed Committee

From 7 men and 6 women, a committee of 5 is to be chosen such that it contains at least 3 men. Find the number of ways.

Solution:
Case 1: 3 men, 2 women → 7C3 × 6C2 = 35 × 15 = 525
Case 2: 4 men, 1 woman → 7C4 × 6C1 = 35 × 6 = 210
Case 3: 5 men, 0 women → 7C5 = 21
Total = 525 + 210 + 21 = 756

Example 7: Divisible Numbers

How many 4-digit numbers divisible by 5 can be formed using digits 0–9 without repetition, order not considered?

Solution:
We just need to select 4 digits out of 10 where the last digit is 0 or 5.
Case 1: Last digit = 0 → choose 3 from 9 → 9C3 = 84
Case 2: Last digit = 5 → choose 3 from 9 → 9C3 = 84
Total = 168

Example 8: Probability Connection

If 6 coins are tossed, what is the probability of getting exactly 4 heads?

Solution:
Total outcomes = 26 = 64
Favorable = 6C4 = 15
Probability = 15 / 64

Relation to Binomial Theorem

The coefficients in the expansion of (a + b)n are given by combinations:

(a+b)n=∑r=0nnCr⋅an−rbr(a + b)n = \sum_{r=0{n} nCr \cdot a{n-r} b^r(a+b)n=r=0∑n​nCr⋅an−rbr

Thus, nCr directly links to binomial coefficients.

Marks Distribution in Exams

• Class 11 Boards: 4–6 marks (committee and card problems are common).
  
• JEE Main: Almost every year 1 question (4 marks) based on combinations, often linked to probability.
  
• KCET/COMEDK: 1–2 direct questions worth 1 mark each.
  
• JEE Advanced: Complex problems involving restrictions and combinations with permutations.
  

Common Mistakes Students Make

1. Confusing with permutations → Remember, in combinations, order does not matter.
2. Forgetting restrictions (like at least one boy, or last digit divisible by 5).
3. Calculation mistakes with factorials in large nCr values.
4. Not using symmetry property (nCr = nC(n-r)) to simplify calculations.

Practice Problems

1. How many ways can 4 cards be chosen from a deck of 52?
2. In how many ways can 6 people be selected from a group of 10?
3. How many different committees of 4 people can be chosen from 7 men and 5 women if the committee must contain exactly 2 women?
4. How many subsets of a set with 12 elements contain 5 elements?
5. A bag has 12 balls. In how many ways can 3 balls be chosen?

FAQs

Q1. What is the difference between permutation and combination?
Permutation → order matters, Combination → order does not matter.

Q2. Why is nCr = nC(n-r)?
Because selecting r objects out of n automatically leaves (n–r) unselected.

Q3. How are combinations used in probability?
They count favorable outcomes when order doesn’t matter.

Q4. Do I need to memorize formulas?
Yes, but more importantly, practice applying them in restricted cases.