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

• Difficulty Level : Hard
• Last Updated : 25 Jun, 2021

Given three numbers n, r and p, compute the 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 solve it on “PRACTICE” first, before moving on to the solution.

We have discussed the 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 the implementation of the above algorithm. In the following implementation, an array fac[] is used to store all the computed factorial values.

C++

 // A modular inverse based solution to// compute nCr % p#include using namespace std; /* Iterative Function to calculate (x^y)%pin O(log y) */unsigned long long power(unsigned long long x,                                  int y, int p){    unsigned long long 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 punsigned long long modInverse(unsigned long long n,                                             int p){    return power(n, p - 2, p);} // Returns nCr % p using Fermat's little// theorem.unsigned long long nCrModPFermat(unsigned long long n,                                 int r, int p){    // If n

Java

 // A modular inverse based solution to// compute nCr %import java.io.*; class GFG {     /* Iterative Function to calculate    (x^y)%p in O(log y) */    static int power(int x, int y, int p)    {         // Initialize result        int res = 1;         // Update x if it is more than or        // equal to p        x = x % p;         while (y > 0) {             // If y is odd, multiply x            // with result            if (y % 2 == 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    static int modInverse(int n, int p)    {        return power(n, p - 2, p);    }     // Returns nCr % p using Fermat's    // little theorem.    static int nCrModPFermat(int n, int r,                             int p)    {           if (n

Python3

 # Python3 function to# calculate nCr % pdef ncr(n, r, p):    # initialize numerator    # and denominator    num = den = 1    for i in range(r):        num = (num * (n - i)) % p        den = (den * (i + 1)) % p    return (num * pow(den,            p - 2, p)) % p # p must be a prime# greater than nn, r, p = 10, 11, 13print("Value of nCr % p is",               ncr(n, r, p))

C#

 // A modular inverse based solution to// compute nCr % pusing System; class GFG {     /* Iterative Function to calculate    (x^y)%p in O(log y) */    static int power(int x, int y, int p)    {         // Initialize result        int res = 1;         // Update x if it is more than or        // equal to p        x = x % p;         while (y > 0) {             // If y is odd, multiply x            // with result            if (y % 2 == 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    static int modInverse(int n, int p)    {        return power(n, p - 2, p);    }     // Returns nCr % p using Fermat's    // little theorem.    static int nCrModPFermat(int n, int r,                             int p)    {       if (n

PHP

 0)    {                 // If y is odd,        // multiply x        // with result        if (\$y & 1)            \$res = (\$res * \$x) % \$p;         // y must be        // even now        // y = y/2        \$y = \$y >> 1;        \$x = (\$x * \$x) % \$p;    }    return \$res;} // Returns n^(-1) mod pfunction modInverse(\$n, \$p){    return power(\$n, \$p - 2, \$p);} // Returns nCr % p using// Fermat's little// theorem.function nCrModPFermat(\$n, \$r, \$p){         if (\$n<\$r)          return 0;      // Base case    if (\$r==0)        return 1;     // Fill factorial array so that we    // can find all factorial of r, n    // and n-r    //\$fac[\$n+1];    \$fac = 1;    for (\$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 Code    // p must be a prime    // greater than n.    \$n = 10;    \$r = 2;    \$p = 13;    echo "Value of nCr % p is ",         nCrModPFermat(\$n, \$r, \$p);         // This code is contributed by Ajit.?>

Javascript


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 write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.