Question

A question paper contains 3 parts A, B and C. There are 5 questions in A, 4 in B and 3 in C. At least 3 questions from A, at least 2 from B and at least 1 from C must be attempted. If a student attempts 7 questions satisfying these conditions, the number of ways is

• A. 330
• B. 402
• C. 390
• D. 430 \bigskip

Hint:

Split selection problems using equations and constraints first; then apply combinations case-by-case.

Answer 1

The Correct Option is C

Concept: This is a partition-based combination problem. We distribute total selections under constraints and then apply combinations.

Step 1: Set up possible distributions.
Let selections from A, B, C be: \[ a+b+c=7 \] with constraints: \[ a\ge3,\quad b\ge2,\quad c\ge1. \] Possible cases: \[ (a,b,c) = (3,3,1),(3,2,2),(4,2,1),(3,4,0 \text{ invalid}),\ldots \] So valid cases: - (3,3,1) - (3,2,2) - (4,2,1)

Step 2: Compute each case.
Case 1: (3,3,1) \[ \binom{5}{3}\binom{4}{3}\binom{3}{1} =10 \cdot 4 \cdot 3 = 120 \] Case 2: (3,2,2) \[ \binom{5}{3}\binom{4}{2}\binom{3}{2} =10 \cdot 6 \cdot 3 = 180 \] Case 3: (4,2,1) \[ \binom{5}{4}\binom{4}{2}\binom{3}{1} =5 \cdot 6 \cdot 3 = 90 \]

Step 3: Add all cases.
\[ 120+180+90=390 \] \[ \boxed{390} \] Hence correct option: \[ \boxed{(C)}. \]