# Compute nCr % p | Set 3 (Using Fermat Little Theorem)

Given three numbers n, r and p, compute value of nCr mod p. Here p is a prime number greater than n. Here nCr is Binomial Coefficient.

Example:

```Input:  n = 10, r = 2, p = 13
Output: 6
Explanation: 10C2 is 45 and 45 % 13 is 6.

Input:  n = 6, r = 2, p = 13
Output: 2
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

We have discussed following methods in previous posts.
Compute nCr % p | Set 1 (Introduction and Dynamic Programming Solution)
Compute nCr % p | Set 2 (Lucas Theorem)

In this post, Fermat Theorem based solution is discussed.

Background:

Fermat’s little theorem and modular inverse
Fermat’s little theorem states that if p is a prime number, then for any integer a, the number ap – a is an integer multiple of p. In the notation of modular arithmetic, this is expressed as:
ap = a (mod p)
For example, if a = 2 and p = 7, 27 = 128, and 128 – 2 = 7 × 18 is an integer multiple of 7.

If a is not divisible by p, Fermat’s little theorem is equivalent to the statement a p – 1 – 1 is an integer multiple of p, i.e
ap-1 = 1 (mod p)

If we multiply both sides by a-1, we get.
ap-2 = a-1 (mod p)

So we can find modular inverse as p-2.

Computation:

```We know the formula for  nCr
nCr = fact(n) / (fact(r) x fact(n-r))
Here fact() means factorial.

nCr % p = (fac[n]* modIverse(fac[r]) % p *
modIverse(fac[n-r]) % p) % p;
Here modIverse() means modular inverse under
modulo p.
```

Following is C++ implementation of the above algorithm. In the following implementation, an array fac[] is used to store all the computed factorial values.

```// A modular inverse based solution to
// compute nCr % p
#include<bits/stdc++.h>
using namespace std;

/* Iterative Function to calculate (x^y)%p
in O(log y) */
int power(int x, int y, int p)
{
int res = 1;      // Initialize result

x = x % p;  // Update x if it is more than or
// equal to p

while (y > 0)
{
// If y is odd, multiply x with result
if (y & 1)
res = (res*x) % p;

// y must be even now
y = y>>1; // y = y/2
x = (x*x) % p;
}
return res;
}

// Returns n^(-1) mod p
int modInverse(int n, int p)
{
return power(n, p-2, p);
}

// Returns nCr % p using Fermat's little
// theorem.
int nCrModPFermat(int n, int r, int p)
{
// Base case
if (r==0)
return 1;

// Fill factorial array so that we
// can find all factorial of r, n
// and n-r
int fac[n+1];
fac[0] = 1;
for (int i=1 ; i<=n; i++)
fac[i] = fac[i-1]*i%p;

return (fac[n]* modInverse(fac[r], p) % p *
modInverse(fac[n-r], p) % p) % p;
}

// Driver program
int main()
{
// p must be a prime greater than n.
int n = 10, r = 2, p = 13;
cout << "Value of nCr % p is "
<< nCrModPFermat(n, r, p);
return 0;
}
```

Output:

```Value of nCr % p is 6
```

Improvements:
In competitive programming, we can pre-compute fac[] for given upper limit so that we don’t have to compute it for every test case. We also can use unsigned long long int everywhere to avoid overflows.

This article is contributed by Nikhil Papisetty. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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