Question

In how many ways can 15 students be seated in a row such that the 2 most talkative children never sit together?

• A. \(14! \cdot 14!\)
• B. \(15 \cdot 14!\)
• C. \(14!\)
• D. \(14! \cdot 13\)
• E. \(15!\)

Hint:

Whenever the condition is “not together”, use: \(\text{Total} - \text{Together}\). For “together”, treat them as a block.

Answer 1

The Correct Option is D

Step 1: Total arrangements without restriction.
15 students in a row can be arranged in \(15!\) ways.

Step 2: Arrangements where the 2 talkative sit together.
Consider the 2 talkative children as a single block. Then we have \(14\) entities (the block + other 13 students).
These can be arranged in \(14!\) ways.
Within the block, the 2 talkative children can be arranged in \(2!\) ways.
So, arrangements with them together \(= 14! \times 2\).

Step 3: Arrangements where they are not together.
Subtract: \[ 15! - (14! \times 2) = 14! \times (15 - 2) = 14! \times 13 \]
Final Answer:
\[ \boxed{14! \cdot 13} \]