Question 1
A committee of 5 members is to be formed from 8 men and 6 women. In how many ways can this be done if the committee must have exactly 2 men and 3 women?
Step-by-Step Solution:
Concept: Combination with constraints - selection from multiple groups with specific requirements.
Given:
- Men available: 8
- Women available: 6
- Committee size: 5
- Required: 2 men and 3 women
Strategy: Select from each group independently, then multiply (Multiplication Principle).
Step 1 - Select Men:
Choose 2 men from 8 men = C(8,2)
C(8,2) = 8! / [2! × 6!] = 28
Step 2 - Select Women:
Choose 3 women from 6 women = C(6,3)
C(6,3) = 6! / [3! × 3!] = 20
Step 3 - Apply Multiplication Principle:
Total ways = C(8,2) × C(6,3)
= 28 × 20
= 560
Key Principle: When selecting from different independent groups with specific requirements from each:
- Calculate selections from each group separately
- Multiply the results
Common Error: Don't add the combinations - multiply them! Each selection from one group can be paired with each selection from the other.
Concept: Combination with constraints - selection from multiple groups with specific requirements.
Given:
- Men available: 8
- Women available: 6
- Committee size: 5
- Required: 2 men and 3 women
Strategy: Select from each group independently, then multiply (Multiplication Principle).
Step 1 - Select Men:
Choose 2 men from 8 men = C(8,2)
C(8,2) = 8! / [2! × 6!] = 28
Step 2 - Select Women:
Choose 3 women from 6 women = C(6,3)
C(6,3) = 6! / [3! × 3!] = 20
Step 3 - Apply Multiplication Principle:
Total ways = C(8,2) × C(6,3)
= 28 × 20
= 560
Key Principle: When selecting from different independent groups with specific requirements from each:
- Calculate selections from each group separately
- Multiply the results
Common Error: Don't add the combinations - multiply them! Each selection from one group can be paired with each selection from the other.