The number of ways of arranging 18 boys such that 3 particular boys are always separate is equal to

Options

A. `18!-16!3!`
B. `16!^(17)P_(3)`
C. `15!^(16)P_(13)`
D. `17!^(17)P_(3)`

Answer

To solve the problem of arranging 18 boys such that 3 particular boys are always separated, we can follow these steps:

### Step 1: Calculate the total arrangements without restrictions
First, we calculate the total number of arrangements of 18 boys without any restrictions. This is given by the factorial of the number of boys:

\[ \text{Total arrangements} = 18! \]

### Step 2: Treat the 3 particular boys as a single unit
Next, to ensure that the 3 particular boys (let's call them A, B, and C) are always separated, we can use the complementary counting method. We first calculate the arrangements where these 3 boys are together.

We can treat A, B, and C as a single unit or block. Thus, instead of 18 boys, we now have 16 units to arrange (the block of A, B, C and the other 15 boys). The number of arrangements of these 16 units is:

\[ \text{Arrangements with A, B, C together} = 16! \]

### Step 3: Arrange the boys within the block
Within the block of A, B, and C, these 3 boys can be arranged among themselves in:

\[ \text{Arrangements of A, B, C} = 3! \]

### Step 4: Calculate the total arrangements where A, B, and C are together
Now, we can combine these two results to find the total arrangements where A, B, and C are together:

\[ \text{Total arrangements with A, B, C together} = 16! \times 3! \]

### Step 5: Use complementary counting
To find the arrangements where A, B, and C are separated, we subtract the arrangements where they are together from the total arrangements:

\[ \text{Arrangements where A, B, C are separated} = 18! - (16! \times 3!) \]

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

\[ \text{Arrangements where A, B, C are separated} = 18! - (6 \times 16!) \]

We can factor out \(16!\):

\[ = 16! \times (18 - 6) = 16! \times 12 \]

### Final Answer
Thus, the number of ways of arranging 18 boys such that 3 particular boys are always separate is:

\[ 12 \times 16! \]