<p>There are 10 intermediary stations between two junctions where an express train stops. If 6 persons board the train at some intermediary station or other during the journey and each of the 6 passengers hold a different variety of ticket of the same class, then find the number of ways in which they can hold their tickets.</p>
Step-by-Step Solution
Key Concept: Each of the 6 passengers must board at different intermediary stations (since they hold different variety tickets), and we need to count ordered selections of 6 distinct boarding stations from 10 available stations.
<p><strong>Step 1:</strong> Identify the constraint. We have 6 distinct passengers (each with different ticket variety) and 10 intermediary stations. Each passenger boards at a different station.</p><p><strong>Step 2:</strong> Since passengers are distinguishable and each must board at a different station, we need to select and arrange 6 stations from 10 available stations in a specific order (Passenger 1 at station A, Passenger 2 at station B, etc.).</p><p><strong>Step 3:</strong> This is a permutation problem: P(10,6) = 10!/(10-6)! = 10!/4!</p><p><strong>Step 4:</strong> Calculate: P(10,6) = 10 × 9 × 8 × 7 × 6 × 5 = 151,200</p><p>∴ <strong>Answer: 151,200 ways</strong></p>
Correct Answer: 151