The are `8` events that can be schedules in a week, then
The total number of ways that the schedule has at least one event in each days of the week is

Options

A. `28xx5040`
B. `7!8!`
C. `7!xx(15!)`
D. None of these

Explanation

To solve the problem of scheduling 8 events in a week such that there is at least one event scheduled each day, we can follow these steps:

### Step 1: Understand the Problem
We have 8 events and 7 days in a week. We need to ensure that each day has at least one event scheduled.

### Step 2: Distribute Events
Since we have 7 days and we want at least one event on each day, we can start by assigning 1 event to each of the 6 days. This accounts for 6 events.

### Step 3: Assign Remaining Events
After assigning 1 event to 6 days, we have 2 events left (8 total events - 6 assigned events = 2 remaining events). We can assign these 2 remaining events to any of the 7 days.

### Step 4: Choose the Day for Extra Events
Now, we need to choose which day will have the 2 extra events. We can choose any of the 7 days for this purpose.

### Step 5: Select the Events
Next, we need to select which 2 events out of the 8 will be assigned to the chosen day. The number of ways to choose 2 events from 8 is given by the combination formula \( \binom{n}{r} \), where \( n \) is the total number of events and \( r \) is the number of events to choose. This can be calculated as:
\[ \binom{8}{2} = \frac{8!}{2!(8-2)!} = \frac{8 \times 7}{2 \times 1} = 28 \]

### Step 6: Arrange the Events
Now, we have 7 groups (one for each day) to arrange. The total arrangements of these 7 groups (days) can be calculated using \( 7! \) (factorial of 7).

### Step 7: Combine the Results
The total number of ways to schedule the events such that there is at least one event each day is given by:
\[ \text{Total Ways} = \binom{8}{2} \times 7! = 28 \times 5040 \]

### Final Calculation
Calculating \( 28 \times 5040 \):
\[ 28 \times 5040 = 141120 \]

Thus, the total number of ways to schedule the events is **141120**.