C Program for Number of stopping station problem

The number of stopping stations problem calculates the ways a train can stop at r stations out of n total stations such that no two stopping stations are consecutive. This is a classic combinatorial problem solved using the formula for selecting non-adjacent elements.

Syntax

C(n-r+1, r) = (n-r+1)! / (r! * (n-2*r+1)!)

Problem Explanation

When a train travels from point X to Y with n intermediate stations, we need to find the number of ways to select r stopping stations such that no two selected stations are adjacent. This constraint transforms the problem into selecting r items from n-r+1 possible positions.

The mathematical formula becomes −

C(n-r+1, r) = (n-r+1)! / (r! × (n-2×r+1)!)

Example

Let's implement the solution to calculate the number of valid stopping patterns −

#include <stdio.h>

int factorial(int n) {
    if (n <= 1) return 1;
    int result = 1;
    for (int i = 2; i <= n; i++) {
        result *= i;
    }
    return result;
}

int main() {
    int n = 16, r = 6;
    
    printf("Total stations: %d<br>", n);
    printf("Stopping stations: %d<br>", r);
    
    /* Check if solution exists */
    if (n - r + 1 < r) {
        printf("No possible ways - not enough stations<br>");
        return 0;
    }
    
    /* Calculate C(n-r+1, r) */
    int numerator = factorial(n - r + 1);
    int denominator = factorial(r) * factorial(n - 2 * r + 1);
    
    int ways = numerator / denominator;
    
    printf("Number of ways: %d<br>", ways);
    
    return 0;
}

Total stations: 16
Stopping stations: 6
Number of ways: 462

Optimized Approach

To avoid large factorial calculations, we can use an optimized method −

#include <stdio.h>

int main() {
    int n = 16, r = 6;
    
    printf("Total stations: %d<br>", n);
    printf("Stopping stations: %d<br>", r);
    
    /* Check validity */
    if (n - r + 1 < r) {
        printf("No possible ways<br>");
        return 0;
    }
    
    /* Calculate using optimized method */
    int numerator = 1, denominator = 1;
    
    /* Calculate r! in denominator */
    for (int i = 1; i <= r; i++) {
        denominator *= i;
    }
    
    /* Calculate (n-r+1) to (n-2*r+2) in numerator */
    for (int i = n - r + 1; i > n - 2 * r + 1; i--) {
        numerator *= i;
    }
    
    int ways = numerator / denominator;
    printf("Number of ways: %d<br>", ways);
    
    return 0;
}

Total stations: 16
Stopping stations: 6
Number of ways: 462

Key Points

• The formula is based on the principle that if we select r non-adjacent positions from n positions, it's equivalent to selecting r positions from n-r+1 positions without restriction.
• The condition n - r + 1 >= r ensures that a valid solution exists.
• The optimized approach prevents integer overflow for large factorial values.

Conclusion

The stopping station problem demonstrates the application of combinatorics in constraint-based selection problems. The solution efficiently calculates valid arrangements using the combination formula C(n-r+1, r).