In how many ways can 12 papers be arranged if the best and the worst paper never come together?

Options

A. 12!/2!
B. 12! – 11!
C. (12! – 11!)/2
D. 12! – 2.11!

Answer

To solve the problem of arranging 12 papers such that the best and worst paper never come together, we can follow these steps:

### Step 1: Calculate the total arrangements without restrictions
First, we calculate the total number of ways to arrange 12 papers without any restrictions. This is given by the factorial of the number of papers.

\[ \text{Total arrangements} = 12! \]

### Step 2: Treat the best and worst paper as one unit
Next, we consider the scenario where the best and worst paper are treated as a single unit or block. By doing this, we temporarily ignore the restriction that they cannot be together.

When we treat the best and worst paper as one unit, we effectively have 11 units to arrange (the block of best and worst paper plus the other 10 papers).

\[ \text{Arrangements with best and worst together} = 11! \]

### Step 3: Arrange the best and worst paper within their block
Within the block of the best and worst paper, these two papers can be arranged in 2 ways (best first or worst first).

\[ \text{Arrangements of best and worst within the block} = 2! \]

### Step 4: Calculate the total arrangements with the best and worst paper together
Now, we can find the total arrangements where the best and worst paper are together by multiplying the arrangements of the 11 units by the arrangements of the best and worst paper within their block.

\[ \text{Total arrangements with best and worst together} = 11! \times 2! \]

### Step 5: Calculate the arrangements where the best and worst paper never come together
To find the arrangements where the best and worst paper never come together, we subtract the arrangements where they are together from the total arrangements without restrictions.

\[ \text{Arrangements where best and worst never together} = 12! - (11! \times 2!) \]

### Step 6: Simplify the expression
Now, we can simplify the expression:

\[ = 12! - 2 \times 11! \]

Factoring out \(11!\):

\[ = 11! \times (12 - 2) = 11! \times 10 \]

### Final Answer
Thus, the number of ways to arrange the 12 papers such that the best and worst paper never come together is:

\[ \text{Final Answer} = 10 \times 11! \]

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