Question: Number of ways in which 7 people can occupy 6 seats, 3 seats on each side in a first class railway c...

Question

Number of ways in which 7 people can occupy 6 seats, 3 seats on each side in a first class railway compartment if two specified persons are to be always included and occupy adjacent seats on the same side is _______.

Answer 1

Solution

To solve this problem, we need to determine the number of ways to arrange 7 people in 6 seats under specific conditions.

Let's break down the problem into steps:

1. Understand the Seating Arrangement: There are 6 seats in total, with 3 seats on each side of a railway compartment.
Let's denote the seats on Side 1 as S11, S12, S13 and on Side 2 as S21, S22, S23.

2. Identify the Specified Persons and Conditions: Let the two specified persons be P1 and P2.
The conditions are:

P1 and P2 are always included among the 6 people occupying the seats.
P1 and P2 must occupy adjacent seats on the same side.

Step-by-step Solution:

Step A: Select the 6 people who will occupy the seats. There are 7 people in total. Since P1 and P2 are specified to be always included, they are two of the six people who will occupy the seats.
We need to choose the remaining 4 people from the remaining 5 people (7 total people - 2 specified people = 5 remaining people).
The number of ways to choose these 4 people from 5 is given by the combination formula:
Number of ways to choose 4 people = ⁵C₄ = $\frac{5!}{4!(5-4)!} = \frac{5!}{4!1!} = 5$ ways.

Step B: Arrange the two specified persons (P1 and P2) according to the condition. P1 and P2 must occupy adjacent seats on the same side. Let's list the possible pairs of adjacent seats on each side:

On Side 1: (S11, S12) and (S12, S13) - 2 pairs.
On Side 2: (S21, S22) and (S22, S23) - 2 pairs.

So, there are a total of 2 + 2 = 4 pairs of adjacent seats on the same side.

Once a pair of seats is chosen (e.g., S11 and S12), P1 and P2 can be arranged in these two seats in 2! ways (P1 in S11, P2 in S12 OR P2 in S11, P1 in S12).
Number of ways to seat P1 and P2 = (Number of adjacent pairs) $\times$ (Ways to arrange P1 and P2 within the pair)
= $4 \times 2! = 4 \times 2 = 8$ ways.

Step C: Arrange the remaining 4 chosen people in the remaining seats. After P1 and P2 are seated, there are $6 - 2 = 4$ seats remaining.
Also, there are $6 - 2 = 4$ people remaining (the 4 people chosen in Step A).
These 4 people can be arranged in the 4 remaining seats in 4! ways.
Number of ways to arrange the remaining 4 people = $4! = 4 \times 3 \times 2 \times 1 = 24$ ways.

Step D: Calculate the total number of ways. To find the total number of ways, we multiply the possibilities from each step:
Total number of ways = (Ways to choose 4 additional people) $\times$ (Ways to seat P1 and P2) $\times$ (Ways to arrange the remaining 4 people)
Total number of ways = ⁵C₄ $\times$ (4 $\times$ 2!) $\times$ 4!
Total number of ways = $5 \times 8 \times 24$
Total number of ways = $40 \times 24 = 960$ .

The final answer is $\boxed{960}$ .

Explanation of the solution:

Select the occupants: Two specified people are always included. From the remaining 5 people, choose 4 to fill the 6 seats (⁵C₄ = 5 ways).
Seat the specified pair: Identify all pairs of adjacent seats on the same side (4 pairs). For each pair, the two specified people can be arranged in 2! ways. Total ways to seat the pair = 4 * 2! = 8 ways.
Seat the remaining occupants: The remaining 4 chosen people can be arranged in the remaining 4 seats in 4! ways (24 ways).
Calculate total ways: Multiply the results from these steps: 5 * 8 * 24 = 960.

The final answer is $\boxed{960}$ .