<p>In how many ways can a team of 6 horses be selected out of a stud of 16, so that there shall always be three out of \(A\ B\ C\ A'\ B'\ C'\), but never \(A\ A'\), \(B\ B'\) or \(C\ C'\) together</p>
Step-by-Step Solution
Key Concept: Use complementary counting or direct casework by recognizing that exactly 3 horses from the special group {A, B, C, A', B', C'} must be selected, with the constraint that no conflicting pair (A,A'), (B,B'), or (C,C') can both be chosen. This is equivalent to selecting 3 non-conflicting horses from 3 pairs, then filling remaining 3 spots from 10 ordinary horses.
<p><strong>Step 1:</strong> Identify that we need exactly 3 horses from {A, B, C, A', B', C'} with no conflicting pair, then 3 from the remaining 10 ordinary horses.</p><p><strong>Step 2:</strong> Count valid selections of 3 from the special group. We have 3 pairs, and must choose 3 horses such that no pair is complete.</p><p>- Choose all 3 from {A, B, C}: C(3,3) = 1 way</p><p>- Choose 2 from {A, B, C} and 1 from their complements: C(3,2) × 2 = 3 × 2 = 6 ways (for each of 2 chosen from {A,B,C}, pick the corresponding prime or non-prime from the other pair)</p><p>- Choose 1 from {A, B, C} and 2 from complements: C(3,1) × C(3,2) = 3 × 3 = 9 ways</p><p>- Choose 0 from {A, B, C} and 3 from {A', B', C'}: C(3,3) = 1 way</p><p><strong>Step 3:</strong> Total valid selections from special group = 1 + 6 + 9 + 1 = 17 ways.</p><p><strong>Step 4:</strong> Select 3 horses from remaining 10 ordinary horses: C(10,3) = 120 ways.</p><p><strong>Step 5:</strong> Total number of ways = 17 × 120 = 2040</p><p>∴ Answer: A</p>
Correct Answer: A