Seating Arrangement Logic

The problem asks for the number of ways to seat 5 people in a row such that a specific couple always sits together.

Treating the Couple as a Unit

To ensure the couple sits together, we can treat them as a single conceptual unit. Let the 5 people be P1, P2, P3, P4, P5, and let P1 and P2 be the couple.

• Consider the couple {P1, P2} as one block, let's call it 'C'.
• Now, we need to arrange the block 'C' and the remaining 3 people (P3, P4, P5).
• This gives us 4 entities to arrange: {C, P3, P4, P5}.

Calculating Arrangements

The number of ways to arrange these 4 distinct entities in a row is given by the factorial of 4:

$ 4! = 4 \times 3 \times 2 \times 1 = 24 $

Within the couple's block 'C', the two people (P1 and P2) can switch places. They can be seated as (P1, P2) or (P2, P1).

The number of ways the couple can arrange themselves is:

$ 2! = 2 \times 1 = 2 $

Total Number of Ways

To find the total number of ways to seat the 5 people with the couple together, we multiply the number of ways to arrange the entities by the number of ways the couple can arrange themselves internally:

$ \text{Total Ways} = (\text{Ways to arrange 4 entities}) \times (\text{Ways to arrange couple internally}) $

$ \text{Total Ways} = 4! \times 2! = 24 \times 2 = 48 $

Therefore, there are 48 distinct ways to seat the 5 people such that the couple sits next to each other.

Option Analysis

The calculated answer is 48, which corresponds to Option B.