Question

\(N\) different objects can be arranged taken all at a time in

• A. \((N-1)!\) ways
• B. \((N+1)!\) ways
• C. \(N!\) ways
• D. \((2N)!\) ways

Hint:

The number of ways to arrange \(N\) distinct objects taken all at a time is: \[ N! \]

Answer 1

The Correct Option is C

Concept:
Arrangement means permutation. If \(N\) different objects are arranged taken all at a time, then all \(N\) objects are placed in \(N\) positions. The number of arrangements of \(N\) different objects is: \[ N! \]

Step 1: Understand the first position.
For the first position, we can choose any one of the \(N\) objects. So the number of choices is: \[ N \]

Step 2: Understand the second position.
After placing one object, \(N-1\) objects remain. So the second position can be filled in: \[ N-1 \] ways.

Step 3: Continue the process.
The third position can be filled in: \[ N-2 \] ways. Continuing like this, the last position can be filled in: \[ 1 \] way.

Step 4: Multiply all choices.
Total arrangements: \[ N(N-1)(N-2)\cdots 3\cdot2\cdot1 \] This is: \[ N! \]

Step 5: Check the options.
Option (A) \((N-1)!\) is for arranging one fewer object.
Option (B) \((N+1)!\) is too large.
Option (C) \(N!\) is correct.
Option (D) \((2N)!\) is not related to arranging \(N\) objects. Hence, the correct answer is: \[ \boxed{(C)\ N!} \]