In this article, you will learn and get code to find and print nPr (Permutation) and nCr (Combination) in C++ language. Here are the list of programs available in this article:
Before going through these programs, let's understand about the formula used to calculate permutation (nPr) and combination (nCr).
The formula to find the value of nPr is:
nPr = n!/(n-r)!
Note - The symbol (!) indicates factorial.
Note - Permutation (nPr) value shows the number of ways to arrange r things out of n
The formula to find the value of nCr is:
nCr = n!/r!(n-r)!
Note - Combination (nCr) value shows the number of ways to select r things out of n
To find nPr in C++ programming, you have to ask from user to enter the value of n and r. In both the program of calculating nPr and nCr, the only required code is to find factorial of a number. Rest of the things are just a normal things like initializing, dividing, printing etc.
The question is, write a program in C++ to find the value of nPr. Here is its answer:
// C++ Program to Find nPr (Permutation)
// ----codescracker.com----
#include<iostream>
using namespace std;
int main()
{
long long fact=1, numerator, denominator;
int perm, n, r, i=1, sub;
cout<<"Enter the Value of n: ";
cin>>n;
cout<<"Enter the Value of r: ";
cin>>r;
while(i<=n)
{
fact = i*fact;
i++;
}
numerator = fact; // n!
sub = n-r;
fact = 1;
i = 1;
while(i<=sub)
{
fact = i*fact;
i++;
}
denominator = fact; // (n-r)!
perm = numerator/denominator;
cout<<"\nPermutation (nPr) = "<<perm;
cout<<endl;
return 0;
}
This program was build and run under Code::Blocks IDE. Here is it sample run:
Now type a number say 5 as value of n, and then type another number say 3 as value of r.
Now press ENTER key to find and print the value of nPr as shown in the output given below:
Note - There are 60 ways to arrange 3 things out of 5.
The following block of code:
while(i<=n)
{
fact = i*fact;
i++;
}
is used to find factorial of n. So the statement:
numerator=fact;
is used to initialized the factorial of n to numerator. That is, numerator = n!
In above program, the factorial of n and n-r gets calculated and then using the formula of nPr, its value gets calculated and printed on output. You can refer to, factorial program for its in-depth detail.
This program asks from user to enter the value of n and r. Based on the values, program finds and prints the value of nCr that tells the number of ways to select r things out of n:
// C++ Program to Find nCr (Combination)
// ----codescracker.com----
#include<iostream>
using namespace std;
int main()
{
long long fact=1, numerator, denominator;
int comb, n, r, i=1, sub;
cout<<"Enter the Value of n: ";
cin>>n;
cout<<"Enter the Value of r: ";
cin>>r;
while(i<=n)
{
fact = i*fact;
i++;
}
numerator = fact; // n!
sub = n-r;
fact = 1;
i = 1;
while(i<=sub)
{
fact = i*fact;
i++;
}
denominator = fact; // (n-r)!
fact = 1;
i = 1;
while(i<=r)
{
fact = i*fact;
i++;
}
denominator = (fact*denominator);
comb = numerator/denominator;
cout<<"\nCombination (nCr) = "<<comb;
cout<<endl;
return 0;
}
Here is its sample run with user input, 5 as value of n and 3 as value of r:
Note - There are 10 ways to select 3 things out of 5.
This program receives the value of n and r. Then calculates and prints permutation and combination value both.
// C++ Program to Find nPr and nCr
// ----codescracker.com----
#include<iostream>
using namespace std;
int main()
{
long long fact=1, numerator, denominator;
int npr, ncr, n, r, i=1, sub;
cout<<"Enter the Value of n: ";
cin>>n;
cout<<"Enter the Value of r: ";
cin>>r;
while(i<=n)
{
fact = i*fact;
i++;
}
numerator=fact; // fact = n!
sub = n-r;
fact=1, i=1;
while(i<=sub)
{
fact = i*fact;
i++;
}
denominator = fact; // fact = (n-r)!
npr = numerator/denominator;
i=1, fact=1;
while(i<=r)
{
fact = i*fact;
i++;
}
ncr = npr/fact; // fact = r!
cout<<"\nPermutation (nPr) = "<<npr;
cout<<"\nCombination (nCr) = "<<ncr;
cout<<endl;
return 0;
}
Here is its sample run with same user input as provided in above program's sample run:
This program uses for loop (instead of while) to do the same job as of previous program.
// Find nPr and nCr using for Loop
// ----codescracker.com----
#include<iostream>
using namespace std;
int main()
{
long long fact, numerator, denominator;
int npr, ncr, n, r, i, sub;
cout<<"Enter the Value of n: ";
cin>>n;
cout<<"Enter the Value of r: ";
cin>>r;
for(i=1, fact=1; i<=n; i++)
fact = i*fact;
numerator=fact; // fact = n!
sub = n-r;
for(i=1, fact=1; i<=sub; i++)
fact = i*fact;
denominator = fact; // fact = (n-r)!
npr = numerator/denominator;
for(i=1, fact=1; i<=r; i++)
fact = i*fact;
ncr = npr/fact; // fact = r!
cout<<"\nPermutation (nPr) = "<<npr;
cout<<"\nCombination (nCr) = "<<ncr;
cout<<endl;
return 0;
}
This program produces the same output as of previous program.
This is the last program of this article, uses some user-defined functions to do the same task as of previous program.
// Find nPr and nCr using Function
// ----codescracker.com----
#include<iostream>
using namespace std;
long long findFact(int);
int findNPR(int, int);
int findNCR(int, int);
int main()
{
int npr, ncr, n, r;
cout<<"Enter the Value of n: ";
cin>>n;
cout<<"Enter the Value of r: ";
cin>>r;
npr = findNPR(n, r);
ncr = findNCR(n, r);
cout<<"\nPermutation (nPr) = "<<npr;
cout<<"\nCombination (nCr) = "<<ncr;
cout<<endl;
return 0;
}
long long findFact(int num)
{
int i=1, fact=1;
while(i<=num)
{
fact = i*fact;
i++;
}
return fact;
}
int findNPR(int n, int r)
{
long long numerator, denominator;
numerator = findFact(n);
denominator = findFact(n-r);
return (numerator/denominator);
}
int findNCR(int n, int r)
{
int npr, ncr;
npr = findNPR(n, r);
ncr = npr/findFact(r);
return ncr;
}