A team of 5 aerospace engineers is to be selected from a pool of 6 aerodynamicists and 4 structural analysts. In how many distinct ways can the team be formed such that it contains at least 3 aerodynamicists?

(A) 186

(B) 120

(C) 210

(D) 246

Answer Key:

Verified in official key

### 1. Strategy & Theoretical Principles The selection requires a 5-member team from 6 aerodynamicists ($A$) and 4 structural analysts ($S$) with the constraint: number of aerodynamicists $k_A \ge 3$. The mutually exclusive valid cases are: - Case 1: Exactly 3 $A$ and 2 $S$ - Case 2: Exactly 4 $A$ and 1 $S$ - Case 3: Exactly 5 $A$ and 0 $S$ ### 2. Analytical Formulation $$\text{Total Ways} = \binom{6}{3}\binom{4}{2} + \binom{6}{4}\binom{4}{1} + \binom{6}{5}\binom{4}{0}$$ ### 3. Step-by-Step Calculation - Case 1: $\binom{6}{3} \times \binom{4}{2} = \frac{6 \times 5 \times 4}{6} \times \frac{4 \times 3}{2} = 20 \times 6 = 120$ - Case 2: $\binom{6}{4} \times \binom{4}{1} = \binom{6}{2} \times 4 = 15 \times 4 = 60$ - Case 3: $\binom{6}{5} \times \binom{4}{0} = 6 \times 1 = 6$ Sum of all cases: $$\text{Total Ways} = 120 + 60 + 6 = 186$$ ### 4. Distractor Forensics - **Option A (186):** Correct combination sum. - **Option B (120):** Considers only the single case of exactly 3 aerodynamicists, ignoring 4 and 5. - **Option C (210):** Erroneous combinatorial multiplication. - **Option D (246):** Total selections without restrictions ($\binom{10}{5} = 252$) minus partial terms. ### 5. Exam Trap & Standard Reference - **Trap Alert:** "At least" means summing across all qualifying cases ($k=3, 4, 5$). - **Standard Reference:** GATE General Aptitude Official Syllabus - Combinatorics.

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