3. The number of ways of arranging r objects chosen from n, where order is important (these are called permutations) is: perm(r, n) = = n! (n-r)! Also, the number of ways of choosing r objects from n without regard to order (these are called combinations) is: n! (n-r)!r! comb(r, n) = Write three functions, fact (n) (= n!), perm(r, n), comb (r, n), and a main program to test them. The main prompts the user to enter the values of r and n then prints the values of r, n, perm(r, n) and comb(r, n) to an output file, permcombout.txt. Note that you must write your functions so that the function perm() calls the function fact () and the function comb () calls both perm() and fact(). Here is a sample input/output: Enter r and n: 3 5 The number of permutations of 3 objects chosen from 5 is 60 The number of combinations of 3 objects chosen from 5 is 10 For marking purposes run your program with r = 4 and n = 8 2

Database System Concepts
7th Edition
ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Chapter1: Introduction
Section: Chapter Questions
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Use c programming for the following question. Complete the coding and plz don't use any other libraries
3. The number of ways of arranging r objects chosen from n, where order is important (these are
called permutations) is:
perm(r, n) =
n!
(n-r)!
Also, the number of ways of choosing r objects from n without regard to order (these are called
combinations) is:
comb(r, n) =
n!
(n − r)!r!
Write three functions, fact (n) (= n!), perm (r, n), comb (r, n), and a main program to
test them. The main prompts the user to enter the values of r and n then prints the values of
r, n, perm(r, n) and comb(r, n) to an output file, permcombout.txt.
Note that you must write your functions so that the function perm() calls the
function fact () and the function comb () calls both perm() and fact().
Here is a sample input/output:
Enter r and n: 35
The number of permutations of 3 objects chosen from 5 is 60
The number of combinations of 3 objects chosen from 5 is 10
For marking purposes run your program with r = 4 and n = 8
2
Transcribed Image Text:3. The number of ways of arranging r objects chosen from n, where order is important (these are called permutations) is: perm(r, n) = n! (n-r)! Also, the number of ways of choosing r objects from n without regard to order (these are called combinations) is: comb(r, n) = n! (n − r)!r! Write three functions, fact (n) (= n!), perm (r, n), comb (r, n), and a main program to test them. The main prompts the user to enter the values of r and n then prints the values of r, n, perm(r, n) and comb(r, n) to an output file, permcombout.txt. Note that you must write your functions so that the function perm() calls the function fact () and the function comb () calls both perm() and fact(). Here is a sample input/output: Enter r and n: 35 The number of permutations of 3 objects chosen from 5 is 60 The number of combinations of 3 objects chosen from 5 is 10 For marking purposes run your program with r = 4 and n = 8 2
#include<stdio.h>
int fact(int x);
int perm(int r, int n);
int comb (int r, int n);
int
main(void) {
int r, n, f, P, C;
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
p=perm (r,n);
c=comb (r,n);
return e;
int fact(int x){
for () {
}
}
int pern (int r, int n) {
}
int comb (int r, int n) {
int c;
c=perm (n,r)/fact(r);
return c;
}
Transcribed Image Text:#include<stdio.h> int fact(int x); int perm(int r, int n); int comb (int r, int n); int main(void) { int r, n, f, P, C; 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 p=perm (r,n); c=comb (r,n); return e; int fact(int x){ for () { } } int pern (int r, int n) { } int comb (int r, int n) { int c; c=perm (n,r)/fact(r); return c; }
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