Permutations & Combinations
Distribution into groups
Grade 11
Question:
<p>In how many ways can 17 persons depart from railway station in 2 cars and 3 autos, given that 2 particular persons depart by the same car (4 persons can sit in a car and 3 persons can sit in an auto)?</p>
<p>\(\dfrac{15!}{2!4!(3!)^3}\)</p>
<p>\(\dfrac{16!}{(2!)^2 4!(3!)^3}\)</p>
<p>\(\dfrac{17!}{2!4!(3!)^3}\)</p>
<p>\(\dfrac{15!}{4!(3!)^3}\)</p>
Step-by-Step Solution
Key Concept: First assign the 2 particular persons to the same car, then distribute remaining 15 persons into the remaining capacities (4+3+3+3=13 available seats), accounting for vehicle arrangements and internal seating orders.
<p><strong>Step 1:</strong> Place the 2 particular persons in the same car. Choose 1 car from 2 cars: <strong>2 ways</strong>. Choose 2 seats from 4 in that car: <strong>P(4,2) = 12 ways</strong>. Total: <strong>2 × 12 = 24 ways</strong>.</p><p><strong>Step 2:</strong> Remaining 15 persons fill remaining seats: 2 seats in car 1 (where 2 particular persons are), 4 seats in car 2, 3 seats in auto 1, 3 seats in auto 2, 3 seats in auto 3. Total = 2+4+3+3+3 = 15 seats available (perfect fit).</p><p><strong>Step 3:</strong> Arrange 15 remaining persons in these 15 distinguishable seats: <strong>P(15,15) = 15! ways</strong>.</p><p><strong>Step 4:</strong> Total arrangements = 24 × 15! = <strong>24 × 15!</strong></p><p><strong>Final Answer:</strong> <strong>24 × 15! = 2 × 3 × 15! = A</strong></p>
Correct Answer: A