Wanted to know how this program functions.
tried hard to figure out the code,but couldn't.
can someone explain to me step by step?

   FIGURE 5-5 Program to compute the combinations function C(n, k)
/*
* File: Combinations.java
* -----------------------
* This program computes the mathematical combinations function
* C(n, k), which is the number of ways of selecting k objects
* from a set of n distinct objects.
*/
import acm.program.*;
public class Combinations extends ConsoleProgram {
/* Runs the program */
public void run() {
int n = readInt("Enter number of objects in the set (n): ");
int k = readInt("Enter number to be chosen (k): ");
println("C(" + n + ", " + k + ") = " + combinations(n, k));
}
/*
* Returns the mathematical combinations function C(n, k),
* which is the number of ways of selecting k objects
* from a set of n distinct objects.
*/
private int combinations(int n, int k) {
return factorial(n) / (factorial(k) * factorial(n - k));
}
/*
* Returns the factorial of n, which is defined as the
* product of all integers from 1 up to n.
*/
private int factorial(int n) {
int result = 1;
for (int i = 1; i <= n; i++) {
result *= i;
}
return result;
}
}             

 

 

Where did you get stuck? How did you try to "figure out the code"? If you haven't already, try stepping through it with a debugger. Or (arguably) better yet, step through it yourself using a piece of paper and a pencil to keep track of what's happening.

 

 

What part of it do you not understand?

That was combination

That was factorial

 

 

-I cant understand where factoria(k) comes from.
-can I write factorial(n-k) as factorial(n)-factorial(k)?

 

 

This isn't a Java related problem, but a math one. factorial(k) is k!, meaning

k * (k - 1) * (k - 2) * (k - 3) ... * 1

No, as for example (4 - 2)! is 2! (equals 2), which is not the same as 4! - 2! (equals 22 if I did my math right)

 

 

No need to calculate all those factorials, most of them cancel out anyway; have a look at this:

public static int comb(int n, int k) {
		
	int m= Math.min(n-k, k);
		
	int c= 1;
	for (int i= 1; i <= m; i++) {
		c*= n--; c/= i;
	}
		
	return c;
}

It takes longer (larger values of n and k) before this method overflows on average.

kind regards,

Jos