In how many ways, `20` guests and `1` host can be arranged around a circular table so that two particular guests are on either side of the host?

Options

A. 18!
B. `18!xx2!`
C. `(18!)/(2!)`
D. 19!

Explanation

To solve the problem of arranging 20 guests and 1 host around a circular table such that two particular guests (let's call them P1 and P2) are on either side of the host (H), we can follow these steps:

### Step 1: Grouping the Host and Two Guests
We can consider the two particular guests (P1 and P2) and the host (H) as a single unit or group. This group will look like either (P1, H, P2) or (P2, H, P1).

### Step 2: Arranging the Group
Since there are two arrangements for this group (P1 on one side of H and P2 on the other, or vice versa), we can arrange this group in \(2!\) (factorial of 2) ways.

### Step 3: Counting Remaining Guests
After forming the group of (P1, H, P2), we have 18 remaining guests (since we started with 20 guests and 1 host, and 3 of them are now grouped together).

### Step 4: Arranging Around a Circular Table
In circular permutations, the number of ways to arrange \(n\) objects is \((n-1)!\). Therefore, for the 18 remaining guests, the number of ways to arrange them around the table is \(18!\).

### Step 5: Total Arrangements
Now, we can find the total number of arrangements by multiplying the arrangements of the group by the arrangements of the remaining guests:
\[ \text{Total arrangements} = 2! \times 18! \]

### Final Calculation
Calculating the values:
- \(2! = 2\)
- Thus, the total arrangements = \(2 \times 18!\)

### Conclusion
The final answer for the number of ways to arrange 20 guests and 1 host around a circular table such that two particular guests are on either side of the host is:
\[ 2 \times 18! \]

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