Introduction
In combinatorics, when we select items from a group where the order of selection does not matter, we use combinations. This problem is a classic example of counting selections under specific constraints.
The Problem Statement
We have an examination paper with 12 questions divided into two parts:
- Part A: 7 questions
- Part B: 5 questions (since $12 - 7 = 5$)
- Task: Select 8 questions in total.
- Constraint: Must select at least 3 questions from each part.
Step-by-Step Solution
To solve this, we define the possible cases that satisfy the requirement of picking at least 3 from each part while totaling 8 selections.
Let $a$ be the number of questions from Part A and $b$ be the number from Part B. We need $a + b = 8$, with $a \geq 3$ and $b \geq 3$.
Identifying Possible Cases:
- Case 1: $a=3, b=5$ (Valid since $a, b \geq 3$)
- Case 2: $a=4, b=4$ (Valid since $a, b \geq 3$)
- Case 3: $a=5, b=3$ (Valid since $a, b \geq 3$)
(Note: $a=6, b=2$ is invalid because $b < 3$; $a=2, b=6$ is invalid because $a < 3$)
Calculations using $^nC_r = \frac{n!}{r!(n-r)!}$:
Case 1 ($a=3, b=5$): $^7C_3 \times ^5C_5 = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} \times 1 = 35 \times 1 = 35$
Case 2 ($a=4, b=4$): $^7C_4 \times ^5C_4 = ^7C_3 \times 5 = 35 \times 5 = 175$
Case 3 ($a=5, b=3$): $^7C_5 \times ^5C_3 = ^7C_2 \times 10 = \frac{7 \times 6}{2} \times 10 = 21 \times 10 = 210$
Final Total:
Total ways = $35 + 175 + 210 = 420$.
Conclusion
By breaking complex constraints into disjoint cases, we can easily calculate total possibilities using the addition principle.