 Hưởng ứng phong trào trồng cây, công ty X tiến
 hành trồng bốn loại cây gồm lim, lộc vừng, bằng lăng và xà cừ trong 7 ngày liên
 tiếp. Trong đó, tổ I phụ tách trồng cây lim và lộc vừng, tổ II phụ trách trồng cây
 bằng lăng và xà cừ. Hỏi có bao nhiêu cách phân chia các ngày để hai tổ trồng cây
 biết rằng mỗi ngày chỉ được trồng một trong bốn loại cây với duy nhất một tổ làm
 việc và không có tổ nào làm việc ba ngày liên tiếp ?

To solve this problem, we need to determine the number of ways to assign the seven days to two teams (Team I and Team II) such that:

1. Each team plants only their assigned trees: Team I plants lim and lộc vừng, and Team II plants bằng lăng and xà cừ.
2. Each day, only one team plants trees.
3. No team works for three consecutive days.

### Step-by-step Solution:

1. **Identify the constraints and variables**:
   - There are 7 days to assign to two teams.
   - Team I and Team II can each work on some of these days, but neither can work for three consecutive days.

2. **Calculate the total unrestricted combinations**:
   Without any restrictions, each of the 7 days can be assigned to either Team I or Team II. This gives a total of:
   \[
   2^7 = 128 \text{ ways}
   \]

3. **Apply the restriction (no team works three days consecutively)**:
   To solve this part, we can use dynamic programming or recursive methods to count valid sequences.

4. **Use recursive relation**:
   Let \(a_n\) represent the number of valid sequences of length \(n\) such that no team works for three consecutive days. Define \(b_n\) as the number of valid sequences of length \(n\) ending in "II" (two consecutive days for Team II), and \(c_n\) as the number of valid sequences of length \(n\) ending in "I" (two consecutive days for Team I).

   We have:
   - \(a_n = a_{n-1} + b_{n-1} + c_{n-1}\)
   - \(b_n = a_{n-1} + c_{n-1}\)
   - \(c_n = a_{n-1} + b_{n-1}\)

   The initial conditions are:
   - \(a_0 = 1\) (One way to have zero days)
   - \(b_0 = 0\)
   - \(c_0 = 0\)

5. **Calculate values up to \(n = 7\)**:
   Start calculating recursively:
   - \(a_1 = a_0 + b_0 + c_0 = 1\)
   - \(b_1 = a_0 + c_0 = 1\)
   - \(c_1 = a_0 + b_0 = 1\)

   Now calculate for \(n = 2\):
   - \(a_2 = a_1 + b_1 + c_1 = 1 + 1 + 1 = 3\)
   - \(b_2 = a_1 + c_1 = 1 + 1 = 2\)
   - \(c_2 = a_1 + b_1 = 1 + 1 = 2\)

   Continue this process up to \(n = 7\):

   - \(a_3 = a_2 + b_2 + c_2 = 3 + 2 + 2 = 7\)
   - \(b_3 = a_2 + c_2 = 3 + 2 = 5\)
   - \(c_3 = a_2 + b_2 = 3 + 2 = 5\)

   - \(a_4 = a_3 + b_3 + c_3 = 7 + 5 + 5 = 17\)
   - \(b_4 = a_3 + c_3 = 7 + 5 = 12\)
   - \(c_4 = a_3 + b_3 = 7 + 5 = 12\)

   - \(a_5 = a_4 + b_4 + c_4 = 17 + 12 + 12 = 41\)
   - \(b_5 = a_4 + c_4 = 17 + 12 = 29\)
   - \(c_5 = a_4 + b_4 = 17 + 12 = 29\)

   - \(a_6 = a_5 + b_5 + c_5 = 41 + 29 + 29 = 99\)
   - \(b_6 = a_5 + c_5 = 41 + 29 = 70\)
   - \(c_6 = a_5 + b_5 = 41 + 29 = 70\)

   - \(a_7 = a_6 + b_6 + c_6 = 99 + 70 + 70 = 239\)

6. **Conclusion**:
   The number of valid ways to assign the 7 days such that no team works three days consecutively is \(a_7 = 239\).

Thus, there are **239 ways** to distribute the 7 days among the two teams under the given conditions.

### Follow-Up Questions:

1. How would the answer change if we had 8 days instead of 7?
2. What if we allow a team to work for up to 4 consecutive days? How many valid distributions would there be then?
3. If one team could work a maximum of 2 consecutive days, how would that affect the total number of valid distributions?
4. What are the implications of increasing the number of days or teams in terms of computational complexity?
5. Can you develop a general formula or algorithm for any number of days \(n\) with the same constraints?

### Tip:

When solving recursive problems, carefully define your base cases and recursive relations. This helps ensure you account for all possibilities and correctly implement constraints.

Discover the mathematical solution to dividing 7 days for planting trees among two teams under specific constraints. Learn about combinatorics and dynamic programming techniques used to solve the problem.

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