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Combinatorics Reading Time: 15 min 8 Problem Types

Standard Selection Problems: The Combinatorics Toolkit

Master committee formations, team selections, and ball picking problems with systematic approaches.

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Why Selection Problems Matter in JEE

Selection problems in combinatorics appear in every JEE paper and test your ability to break down complex word problems into simple combinations and permutations. Mastering these will help you:

  • Solve 3-6 mark problems in under 5 minutes
  • Develop systematic thinking for complex scenarios
  • Avoid common pitfalls like double counting
  • Build conceptual clarity for probability problems

Essential Combinatorics Formulas

Fundamental Formulas

Combination Formula:

$$ C(n, r) = \binom{n}{r} = \frac{n!}{r!(n-r)!} $$

Use when order doesn't matter

Permutation Formula:

$$ P(n, r) = \frac{n!}{(n-r)!} $$

Use when order matters

🎯 Decision Framework

Use Combination when:
  • Selecting committee members
  • Choosing teams without specific roles
  • Picking balls from a bag
Use Permutation when:
  • Assigning specific roles (captain, vice-captain)
  • Arranging people in a line
  • Forming numbers with digits

1. Committee Formation Problems

Basic Committee Easy

Problem 1: Simple Committee Selection

From 10 people, in how many ways can we select a committee of 4 people?

Solution Approach:

Step 1: Identify this as a combination problem (order doesn't matter)

Step 2: Apply combination formula:

$$ C(10, 4) = \frac{10!}{4!(10-4)!} = \frac{10 × 9 × 8 × 7}{4 × 3 × 2 × 1} = 210 $$

Step 3: Verify: No restrictions, no specific roles

Answer: 210 ways

Restricted Committee Medium

Problem 2: Committee with Restrictions

From 7 men and 5 women, select a committee of 5 with at least 3 men.

Solution Approach:

Step 1: Break into cases based on "at least 3 men":

• Case 1: 3 men + 2 women

• Case 2: 4 men + 1 woman

• Case 3: 5 men + 0 women

Step 2: Calculate each case:

Case 1: $C(7,3) × C(5,2) = 35 × 10 = 350$

Case 2: $C(7,4) × C(5,1) = 35 × 5 = 175$

Case 3: $C(7,5) × C(5,0) = 21 × 1 = 21$

Step 3: Add all cases: $350 + 175 + 21 = 546$

Answer: 546 ways

2. Team Formation Problems

Specific Roles Medium

Problem 3: Team with Specific Roles

From 12 players, select a team of 5 with specific roles: captain, vice-captain, and 3 players.

Solution Approach:

Step 1: Select captain and vice-captain (order matters):

$$ P(12, 2) = 12 × 11 = 132 $$

Step 2: Select remaining 3 players from remaining 10 (order doesn't matter):

$$ C(10, 3) = 120 $$

Step 3: Multiply: $132 × 120 = 15,840$

Answer: 15,840 ways

Multiple Teams Hard

Problem 4: Dividing into Multiple Teams

Divide 10 different people into two teams of 5 each. In how many ways?

Solution Approach:

Step 1: Select first team of 5 from 10:

$$ C(10, 5) = 252 $$

Step 2: The remaining 5 automatically form second team

Step 3: Since teams are unlabeled, divide by 2 to avoid double counting:

$$ \frac{252}{2} = 126 $$

Answer: 126 ways

3. Ball Selection Problems

Identical Balls Easy

Problem 5: Selecting Identical Balls

A bag contains 8 identical red balls and 6 identical blue balls. In how many ways can we select 5 balls?

Solution Approach:

Step 1: Since balls are identical, we count distributions:

Step 2: Let r = number of red balls, b = number of blue balls

Step 3: r + b = 5, with 0 ≤ r ≤ 8, 0 ≤ b ≤ 6

Step 4: Possible (r, b) pairs: (5,0), (4,1), (3,2), (2,3), (1,4), (0,5)

Step 5: But (0,5) is invalid since we only have 4 blue balls maximum

Step 6: Valid pairs: (5,0), (4,1), (3,2), (2,3), (1,4)

Answer: 5 ways

Distinct Balls Hard

Problem 6: Selecting Distinct Balls with Conditions

A bag has 5 distinct red balls, 4 distinct blue balls, and 3 distinct green balls. Select 6 balls with at least 2 of each color.

Solution Approach:

Step 1: Start with minimum required: 2 red + 2 blue + 2 green = 6 balls

Step 2: Since we need exactly 6 balls, this is the only possibility

Step 3: Select 2 red from 5: $C(5,2) = 10$

Step 4: Select 2 blue from 4: $C(4,2) = 6$

Step 5: Select 2 green from 3: $C(3,2) = 3$

Step 6: Multiply: $10 × 6 × 3 = 180$

Answer: 180 ways

⚠️ Common Mistakes to Avoid

Conceptual Errors:

  • Using permutation when combination is needed
  • Forgetting to divide by 2 for identical teams
  • Missing "at least" or "at most" conditions
  • Double counting identical objects

Calculation Errors:

  • Not simplifying factorial expressions
  • Missing cases in case-based problems
  • Incorrect application of addition/multiplication principles
  • Confusing C(n,r) with P(n,r) formulas

🔧 Systematic Problem Solving Framework

Step 1: Identify the Type

  • Committee formation → Combination
  • Team with roles → Permutation + Combination
  • Ball selection → Check if objects are identical or distinct

Step 2: Note Restrictions

  • "At least" → Break into cases
  • "At most" → Use complement if easier
  • Specific inclusions/exclusions → Adjust total count

Step 3: Apply Formulas

  • Use combination C(n,r) when order doesn't matter
  • Use permutation P(n,r) when order matters
  • Multiply for independent choices
  • Add for mutually exclusive cases

Step 4: Check for Overcounting

  • Divide by k! if k objects are identical
  • Check if teams/groups are labeled or unlabeled
  • Verify each case is distinct

📝 Practice Problems

Test your understanding with these problems:

1. From 8 men and 7 women, select a committee of 6 with exactly 4 men.

Hint: Use multiplication principle

2. In how many ways can 12 different books be distributed equally among 3 students?

Hint: Consider if students are distinct

3. A box contains 5 red, 4 blue, and 3 green distinct balls. Select 4 balls with at least one of each color.

Hint: Break into cases based on the extra ball

📋 Quick Reference Guide

Key Formulas:

  • $C(n,r) = \frac{n!}{r!(n-r)!}$
  • $P(n,r) = \frac{n!}{(n-r)!}$
  • $C(n,r) = C(n,n-r)$
  • $P(n,r) = r! × C(n,r)$

Common Scenarios:

  • Committee without roles → Combination
  • Team with captain → Permutation × Combination
  • Identical objects → Distribution counting
  • "At least" conditions → Case analysis

Master Combinatorics for JEE Success!

These selection problems are guaranteed in every JEE paper. Practice systematically and build confidence.

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