In an examination, a question paper consists of 12 questions divided into two sections i.e. A and B, containing 7 and 5 questions, respectively. A student is required to attempt 8 questions in all and first question of section A is compulsory. In how many ways can the student sel

Question

In an examination, a question paper consists of 12 questions divided into two sections i.e. A and B, containing 7 and 5 questions, respectively. A student is required to attempt 8 questions in all and first question of section A is compulsory. In how many ways can the student select the questions if at least 3 questions are to be attempted from each section.

Accepted Answer

> [!TLDR] Question 1 of section A is fixed; the remaining 7 picks must be split between section A's other 6 questions and section B's 5 questions so that each section's total (A including Q1, B) is at least 3 — sum the three valid splits.

> [!FORMULA]
> $^nC_r=\dfrac{n!}{r!(n-r)!}$ — number of ways to choose $r$ from $n$. Independent simultaneous choices are multiplied (product rule); mutually exclusive cases are added (sum rule).

1. Q1 of section A is compulsory (already counted as 1 question from A). The student must attempt 8 questions in all, so $8-1=7$ more questions are chosen from the remaining $6$ questions of section A and the $5$ questions of section B.
2. Let $a=$ number of *additional* questions chosen from section A (from its remaining 6) and $b=$ number chosen from section B (from its 5). Then $a+b=7$.
3. The condition "at least 3 from each section" applies to each section's **total**: section A's total is $a+1$ (including the compulsory Q1), so $a+1\ge3\Rightarrow a\ge2$; section B's total is simply $b$, so $b\ge3$.
4. Since $a+b=7$ and $b\ge3$, we get $a\le4$; combined with $a\ge2$, the valid integer values are $a\in\{2,3,4\}$, giving $b\in\{5,4,3\}$ respectively (each within the available $6$ and $5$ questions).
5. **Case $a=2,\,b=5$:** $^6C_2	imes{}^5C_5=15	imes1=15$.
6. **Case $a=3,\,b=4$:** $^6C_3	imes{}^5C_4=20	imes5=100$.
7. **Case $a=4,\,b=3$:** $^6C_4	imes{}^5C_3=15	imes10=150$.
8. These three cases are mutually exclusive, so total ways $=15+100+150=265$.
9. Self-check: $15+100=115$, $115+150=265$ ✓.

> [!ANSWER] $265$ ways.

Concept understanding — Miscellaneous

The Concept of "Miscellaneous" – First Principles

Intuition First

The Precise Statement

Why It Matters

A Concrete Example (from a typical Class 12 Maths chapter)

The Bottom Line

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