Given three values, N, L and R, the task is to calculate the sum of binomial coefficients (nCr) for all values of r from L to R.
Examples:
Input: N = 5, L = 0, R = 3
Output: 26
Explanation: Sum of 5C0 + 5C1 + 5C2 + 5C3 = 1 + 5 + 10 + 10 = 26.Input: N = 3, L = 3, R = 3
Output: 1
Approach(Using factorial function): Solve this problem by straightforward calculating nCr by using the formula n! / (r!(n−r)!) and calculating factorial recursively for every value of r from L to R.
Below is the implementation of the above approach:
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to find the factorial
// of a given number
long long factorial(long long num)
{
if (num == 0 || num == 1)
return 1;
else
return num * factorial(num - 1);
}
// Function to calculate the sum
// of binomial coefficients(nCr) for
// all values of r from L to R
long long sumOfnCr(int n, int R, int L)
{
long long r;
long long res = 0;
for (r = L; r <= R; r++)
res += (factorial(n)
/ (factorial(r)
* factorial(n - r)));
return res;
}
// Driver Code
int main()
{
int N = 5, L = 0, R = 3;
cout << sumOfnCr(N, R, L);
return 0;
}
// JAVA program for the above approach
import java.io.*;
class GFG {
// Function to find the factorial
// of a given number
static long factorial(long num)
{
if (num == 0 || num == 1)
return 1;
else
return num * factorial(num - 1);
}
// Function to calculate the sum
// of binomial coefficients(nCr) for
// all values of r from L to R
static long sumOfnCr(int n, int R, int L)
{
long r;
long res = 0;
for (r = L; r <= R; r++)
res += (factorial(n)
/ (factorial(r) * factorial(n - r)));
return res;
}
// Driver Code
public static void main(String[] args)
{
int N = 5, L = 0, R = 3;
long ans = sumOfnCr(N, R, L);
System.out.println(ans);
}
}
// This code is contributed by Taranpreet
# Python code for the above approach
# Function to find the factorial
# of a given number
def factorial(num):
if (num == 0 or num == 1):
return 1;
else:
return num * factorial(num - 1);
# Function to calculate the sum
# of binomial coefficients(nCr) for
# all values of r from L to R
def sumOfnCr(n, R, L):
res = 0;
for r in range(L, R + 1):
res += (factorial(n) / (factorial(r) * factorial(n - r)));
return res;
# Driver Code
N = 5
L = 0
R = 3;
print((int)(sumOfnCr(N, R, L)))
# This code is contributed by gfgking
// C# program for the above approach
using System;
class GFG {
// Function to find the factorial
// of a given number
static long factorial(long num)
{
if (num == 0 || num == 1)
return 1;
else
return num * factorial(num - 1);
}
// Function to calculate the sum
// of binomial coefficients(nCr) for
// all values of r from L to R
static long sumOfnCr(int n, int R, int L)
{
long r;
long res = 0;
for (r = L; r <= R; r++)
res += (factorial(n)
/ (factorial(r) * factorial(n - r)));
return res;
}
// Driver Code
public static void Main()
{
int N = 5, L = 0, R = 3;
Console.Write(sumOfnCr(N, R, L));
}
}
// This code is contributed by ukasp.
<script>
// JavaScript code for the above approach
// Function to find the factorial
// of a given number
function factorial(num) {
if (num == 0 || num == 1)
return 1;
else
return num * factorial(num - 1);
}
// Function to calculate the sum
// of binomial coefficients(nCr) for
// all values of r from L to R
function sumOfnCr(n, R, L) {
let r;
let res = 0;
for (r = L; r <= R; r++)
res += (factorial(n)
/ (factorial(r)
* factorial(n - r)));
return res;
}
// Driver Code
let N = 5, L = 0, R = 3;
document.write(sumOfnCr(N, R, L));
// This code is contributed by Potta Lokesh
</script>
Output
26
Time Complexity: O(N * (R - L))
Auxiliary Space: O(N)
Approach (Without using factorial function): This approach for finding the sum of binomial coefficients (nCr) for all values of r from L to R can be implemented using two nested loops. The outer loop will iterate from L to R, and the inner loop will calculate the binomial coefficient for each value of r using the formula:
nCr = n! / (r! * (n-r)!)
where n is the given number, r is the current value of the inner loop, and ! denotes the factorial function.
The sum of all binomial coefficients can be accumulated in a variable initialized to zero before the loops start.
Steps to implement the above approach:
- Declare and initialize the variables N, L, and R respectively.
- Call the sumOfnCr function, passing N, R, and L as arguments.
- Within the sumOfnCr function, declare a long long variable named res and initialize it to 0.
- Start a for loop with a variable r, which runs from L to R.
- Within the for loop, declare a long long variable named nCr and initialize it to 1.
- Start another for loop with a variable i, which runs from 1 to r.
- Within the inner for loop, multiply nCr by (n-i+1) and then divide it by i.
- After the inner for loop ends, add nCr to the res variable.
- After the outer for loop ends, return the res variable.
- End the sumOfnCr function.
- Print the result of the sumOfnCr function using cout.
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to calculate the sum
// of binomial coefficients(nCr) for
// all values of r from L to R
long long sumOfnCr(int n, int R, int L)
{
long long res = 0;
for (int r = L; r <= R; r++) {
long long nCr = 1;
for (int i = 1; i <= r; i++) {
nCr *= (n - i + 1);
nCr /= i;
}
res += nCr;
}
return res;
}
// Driver Code
int main()
{
int N = 5, L = 0, R = 3;
cout << sumOfnCr(N, R, L);
return 0;
}
import java.math.BigInteger;
public class BinomialCoefficientSum {
// Function to calculate the sum of binomial coefficients (nCr)
// for all values of r from L to R
public static BigInteger sumOfnCr(int n, int R, int L) {
// Initialize the result as zero
BigInteger res = BigInteger.ZERO;
// Iterate through all values of r from L to R
for (int r = L; r <= R; r++) {
// Initialize nCr as 1
BigInteger nCr = BigInteger.ONE;
// Calculate nCr using multiplicative formula
for (int i = 1; i <= r; i++) {
// Multiplicative formula for nCr: nCr = (n - i + 1) / i * nCr
nCr = nCr.multiply(BigInteger.valueOf(n - i + 1)).divide(BigInteger.valueOf(i));
}
// Add the calculated nCr to the result
res = res.add(nCr);
}
// Return the final result
return res;
}
public static void main(String[] args) {
// Input values
int N = 5, L = 0, R = 3;
// Calculate and print the sum of binomial coefficients
System.out.println(sumOfnCr(N, R, L));
}
}
# Function to calculate the sum
# of binomial coefficients(nCr) for
# all values of r from L to R
def sumOfnCr(n, R, L):
res = 0
for r in range(L, R+1):
nCr = 1
for i in range(1, r+1):
nCr *= (n - i + 1)
nCr //= i
res += nCr
return res
# Driver Code
N = 5
L = 0
R = 3
print(sumOfnCr(N, R, L))
using System;
class Program
{
// Function to calculate the sum of binomial coefficients(nCr) for
// all values of r from L to R
static long SumOfnCr(int n, int R, int L)
{
long res = 0;
for (int r = L; r <= R; r++)
{
long nCr = 1;
for (int i = 1; i <= r; i++)
{
nCr *= (n - i + 1);
nCr /= i;
}
res += nCr;
}
return res;
}
static void Main(string[] args)
{
int N = 5, L = 0, R = 3;
Console.WriteLine(SumOfnCr(N, R, L));
}
}
// This code is contributed by shivamgupta310570
// Javascript code addition
// Function to calculate the sum
// of binomial coefficients(nCr) for
// all values of r from L to R
function sumOfnCr(n, R, L) {
let res = 0;
for (let r = L; r <= R; r++) {
let nCr = 1;
for (let i = 1; i <= r; i++) {
nCr *= (n - i + 1);
nCr /= i;
}
res += nCr;
}
return res;
}
// Driver Code
let N = 5, L = 0, R = 3;
console.log(sumOfnCr(N, R, L));
// The code is contributed by Nidhi goel.
Output
26
Time Complexity: O(N * (R - L))
Auxiliary Space: O(N)
