A train time table must be compiled for various days of the week, so that two trains twice a day depart for three days, one train daily for two days, and three trains once a day for two days. How many different time tables can be compiled? a. 140 b. 210 c. 133 d. 72

Options

A. 140
B. 210
C. 133
D. 72

Answer

To solve the problem of compiling a train timetable for various days of the week, we need to consider the different train schedules provided and how they can be arranged over the week. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Requirements
We have:
- Two trains departing twice a day for three days.
- One train departing daily for two days.
- Three trains departing once a day for two days.

### Step 2: Determine the Total Days
There are a total of 7 days in a week.

### Step 3: Identify the Groups
We can categorize the days based on the train schedules:
- **Group A**: 3 days with 2 trains (2 trains twice a day)
- **Group B**: 2 days with 1 train (1 train daily)
- **Group C**: 2 days with 3 trains (3 trains once a day)

### Step 4: Calculate the Total Arrangements
The total arrangements can be calculated using the formula for permutations of multiset:

\[ \text{Total Arrangements} = \frac{n!}{n_1! \times n_2! \times n_3!} \]

Where:
- \( n \) is the total number of days (7),
- \( n_1 \) is the number of days in Group A (3),
- \( n_2 \) is the number of days in Group B (2),
- \( n_3 \) is the number of days in Group C (2).

### Step 5: Plug in the Values
Using the values we have:

\[ \text{Total Arrangements} = \frac{7!}{3! \times 2! \times 2!} \]

### Step 6: Calculate Factorials
Now, we calculate the factorials:
- \( 7! = 5040 \)
- \( 3! = 6 \)
- \( 2! = 2 \)

### Step 7: Substitute and Simplify
Substituting the factorial values into the equation:

\[ \text{Total Arrangements} = \frac{5040}{6 \times 2 \times 2} = \frac{5040}{24} = 210 \]

### Step 8: Conclusion
Thus, the total number of different timetables that can be compiled is **210**.

### Final Answer
The correct option is **b. 210**.

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