There are 10 stations on a circular path. A train has to stop at 4 stations such that no two stations are adjacent. The number of such selections must be: (a) 25 (b) 35 (c) 210 (d) 50

Options

A. 25
B. 35
C. 210
D. 50

Answer

To solve the problem of selecting 4 stations from 10 arranged in a circular path such that no two selected stations are adjacent, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Problem**: We have 10 stations arranged in a circle, and we need to select 4 stations such that no two selected stations are adjacent.

2. **Transforming the Circular Arrangement**: To make the problem simpler, we can break the circular arrangement into a linear one. We can fix one station (say station 1) and consider the remaining stations (2 to 10) in a linear fashion. This way, we can avoid the adjacency issue that arises from the circular arrangement.

3. **Identifying the Gaps**: When we select 4 stations, we need to ensure that there is at least one unselected station between any two selected stations. If we select 4 stations, we will have 4 gaps created by the selected stations, and we will need to place at least 1 unselected station in each of these gaps.

4. **Calculating Remaining Stations**: Since we have 10 stations and we are selecting 4, we have 6 unselected stations. After placing 1 unselected station in each of the 4 gaps (to ensure no two selected stations are adjacent), we are left with \(6 - 4 = 2\) unselected stations to freely distribute.

5. **Distributing Remaining Stations**: Now, we need to distribute these 2 remaining unselected stations into the 4 gaps. This is a classic "stars and bars" problem where we need to find the number of ways to distribute \(n\) indistinguishable objects (the remaining unselected stations) into \(k\) distinguishable boxes (the gaps).

6. **Applying the Stars and Bars Theorem**: The number of ways to distribute \(n\) indistinguishable objects into \(k\) distinguishable boxes is given by the formula:
\[ \text{Number of ways} = \binom{n + k - 1}{k - 1} \]
In our case, \(n = 2\) (remaining unselected stations) and \(k = 4\) (gaps). Thus, we need to calculate:
\[ \binom{2 + 4 - 1}{4 - 1} = \binom{5}{3} \]

7. **Calculating the Binomial Coefficient**:
\[ \binom{5}{3} = \frac{5!}{3!(5-3)!} = \frac{5 \times 4}{2 \times 1} = 10 \]

8. **Considering the Circular Arrangement**: Since we fixed one station to avoid circular adjacency, we need to multiply the result by the number of ways to select the first station (which can be any of the 10 stations). Thus, the total number of selections is:
\[ 10 \times 10 = 100 \]

9. **Final Adjustment**: However, we need to account for the fact that we have counted each arrangement multiple times due to the circular nature. The correct number of ways to select 4 stations such that no two are adjacent is actually:
\[ \frac{10 \times 10}{4} = 25 \]

### Conclusion:
The number of ways to select 4 stations such that no two are adjacent is **25**.