An engineer is required to visit a factroy for exactly four days during the first 15 days of every month and it I mandatory that no two visits take on consecutive days . Then the number of all possible ways in which such visits to the factory can be made by the engineer during 1- 15 June 2021 is ______

Answer

To solve the problem of how many ways the engineer can visit the factory for exactly 4 days during the first 15 days of June 2021, with the condition that no two visits can occur on consecutive days, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Problem**:
The engineer needs to visit the factory on 4 non-consecutive days out of the first 15 days of the month.

2. **Calculating the Remaining Days**:
If the engineer visits on 4 days, then there are \(15 - 4 = 11\) days left that he cannot visit.

3. **Creating Gaps**:
To ensure that no two visits are on consecutive days, we can think of the days he visits as creating gaps. If we denote the days he visits as V (for visit) and the days he does not visit as N (for no visit), we can represent the situation as follows:
- V N V N V N V
This representation shows that there must be at least one N (no visit) between each V (visit).

4. **Counting Gaps**:
Between the 4 visits, there are 3 mandatory gaps (N) that must exist to separate the visits. Thus, we have:
- 4 visits + 3 gaps = 7 days accounted for.
This leaves us with \(15 - 7 = 8\) days that can be freely assigned as either visits or non-visits.

5. **Total Gaps Available**:
After accounting for the mandatory gaps, we can visualize the situation as having a total of \(11 + 1 = 12\) gaps (the 11 remaining days plus 1 additional gap before the first visit).

6. **Choosing Gaps for Visits**:
We need to choose 4 gaps from these 12 available gaps to place the visits. This can be calculated using the combination formula:
\[ \text{Number of ways} = \binom{12}{4} \]

7. **Calculating the Combination**:
Now we compute \(\binom{12}{4}\):
\[ \binom{12}{4} = \frac{12!}{4!(12-4)!} = \frac{12 \times 11 \times 10 \times 9}{4 \times 3 \times 2 \times 1} = \frac{11880}{24} = 495 \]

8. **Final Answer**:
Therefore, the total number of ways the engineer can visit the factory is **495**.