Sum of Binomial coefficients (original) (raw)
Last Updated : 30 Apr, 2021
Given a positive integer n, the task is to find the sum of binomial coefficient i.e
nC0 + nC1 + nC2 + ....... + nCn-1 + nCn
Examples:
Input : n = 4 Output : 16 4C0 + 4C1 + 4C2 + 4C3 + 4C4 = 1 + 4 + 6 + 4 + 1 = 16
Input : n = 5 Output : 32
Method 1 (Brute Force):
The idea is to evaluate each binomial coefficient term i.e nCr, where 0 <= r <= n and calculate the sum of all the terms.
Below is the implementation of this approach:
C++ `
// CPP Program to find the sum of Binomial // Coefficient. #include <bits/stdc++.h> using namespace std;
// Returns value of Binomial Coefficient Sum int binomialCoeffSum(int n) { int C[n + 1][n + 1];
// Calculate value of Binomial Coefficient
// in bottom up manner
for (int i = 0; i <= n; i++) {
for (int j = 0; j <= min(i, n); j++) {
// Base Cases
if (j == 0 || j == i)
C[i][j] = 1;
// Calculate value using previously
// stored values
else
C[i][j] = C[i - 1][j - 1] + C[i - 1][j];
}
}
// Calculating the sum.
int sum = 0;
for (int i = 0; i <= n; i++)
sum += C[n][i];
return sum;}
/* Driver program to test above function*/ int main() { int n = 4; printf("%d", binomialCoeffSum(n)); return 0; }
Java
// Java Program to find the sum // of Binomial Coefficient.
class GFG {
// Returns value of Binomial
// Coefficient Sum
static int binomialCoeffSum(int n)
{
int C[][] = new int[n + 1][n + 1];
// Calculate value of Binomial
// Coefficient in bottom up manner
for (int i = 0; i <= n; i++)
{
for (int j = 0; j <= Math.min(i, n); j++)
{
// Base Cases
if (j == 0 || j == i)
C[i][j] = 1;
// Calculate value using previously
// stored values
else
C[i][j] = C[i - 1][j - 1] +
C[i - 1][j];
}
}
// Calculating the sum.
int sum = 0;
for (int i = 0; i <= n; i++)
sum += C[n][i];
return sum;
}
/* Driver program to test above function*/
public static void main(String[] args)
{
int n = 4;
System.out.println(binomialCoeffSum(n));
}}
// This code is contributed by prerna saini.
Python3
Python Program to find the sum
of Binomial Coefficient.
import math
Returns value of Binomial
Coefficient Sum
def binomialCoeffSum( n):
C = [[0]*(n+2) for i in range(0,n+2)]
# Calculate value of Binomial
# Coefficient in bottom up manner
for i in range(0,n+1):
for j in range(0, min(i, n)+1):
# Base Cases
if (j == 0 or j == i):
C[i][j] = 1
# Calculate value using previously
# stored values
else:
C[i][j] = C[i - 1][j - 1] + C[i - 1][j]
# Calculating the sum.
sum = 0
for i in range(0,n+1):
sum += C[n][i]
return sum
Driver program to test above function
n = 4 print(binomialCoeffSum(n))
This code is contributed by Gitanjali.
C#
// C# program to find the sum // of Binomial Coefficient. using System;
class GFG {
// Returns value of Binomial
// Coefficient Sum
static int binomialCoeffSum(int n)
{
int[, ] C = new int[n + 1, n + 1];
// Calculate value of Binomial
// Coefficient in bottom up manner
for (int i = 0; i <= n; i++)
{
for (int j = 0; j <= Math.Min(i, n); j++)
{
// Base Cases
if (j == 0 || j == i)
C[i, j] = 1;
// Calculate value using previously
// stored values
else
C[i, j] = C[i - 1, j - 1] + C[i - 1, j];
}
}
// Calculating the sum.
int sum = 0;
for (int i = 0; i <= n; i++)
sum += C[n, i];
return sum;
}
/* Driver program to test above function*/
public static void Main()
{
int n = 4;
Console.WriteLine(binomialCoeffSum(n));
}}
// This code is contributed by vt_m.
PHP
JavaScript
`
Output:
16
Method 2 (Using Formula):
This can be proved in 2 ways.
First Proof: Using Principle of induction.
For basic step, n = 0
LHS = 0C0 = (0!)/(0! * 0!) = 1/1 = 1.
RHS= 20 = 1.
LHS = RHS
For induction step:
Let k be an integer such that k > 0 and for all r, 0 <= r <= k, where r belong to integers,
the formula stand true.
Therefore,
kC0 + kC1 + kC2 + ....... + kCk-1 + kCk = 2k
Now, we have to prove for n = k + 1,
k+1C0 + k+1C1 + k+1C2 + ....... + k+1Ck + k+1Ck+1 = 2k+1
LHS = k+1C0 + k+1C1 + k+1C2 + ....... + k+1Ck + k+1Ck+1
(Using nC0 = 0 and n+1Cr = nCr + nCr-1)
= 1 + kC0 + kC1 + kC1 + kC2 + ...... + kCk-1 + kCk + 1
= kC0 + kC0 + kC1 + kC1 + ...... + kCk-1 + kCk-1 + kCk + kCk
= 2 X ? nCr
= 2 X 2k
= 2k+1
= RHS
Second Proof: Using Binomial theorem expansion
Binomial expansion state,
(x + y)n = nC0 xn y0 + nC1 xn-1 y1 + nC2 xn-2 y2 + ......... + nCn-1 x1 yn-1 + nCn x0 yn
Put x = 1, y = 1
(1 + 1)n = nC0 1n 10 + nC1 xn-1 11 + nC2 1n-2 12 + ......... + nCn-1 11 1n-1 + nCn 10 1n
2n = nC0 + nC1 + nC2 + ....... + nCn-1 + nCn
Below is implementation of this approach:
C++ `
// CPP Program to find sum of Binomial // Coefficient. #include <bits/stdc++.h> using namespace std;
// Returns value of Binomial Coefficient Sum // which is 2 raised to power n. int binomialCoeffSum(int n) { return (1 << n); }
/* Driver program to test above function*/ int main() { int n = 4; printf("%d", binomialCoeffSum(n)); return 0; }
Java
// Java Program to find sum // of Binomial Coefficient. import java.io.*;
class GFG { // Returns value of Binomial // Coefficient Sum which is // 2 raised to power n. static int binomialCoeffSum(int n) { return (1 << n); }
// Driver Code
public static void main (String[] args)
{
int n = 4;
System.out.println(binomialCoeffSum(n));
}}
// This code is contributed // by akt_mit.
Python3
Python Program to find the sum
of Binomial Coefficient.
import math
Returns value of Binomial
Coefficient Sum
def binomialCoeffSum( n):
return (1 << n);Driver program to test
above function
n = 4 print(binomialCoeffSum(n))
This code is contributed
by Gitanjali.
C#
// C# Program to find sum of // Binomial Coefficient. using System;
class GFG {
// Returns value of Binomial Coefficient Sum
// which is 2 raised to power n.
static int binomialCoeffSum(int n)
{
return (1 << n);
}
/* Driver program to test above function*/
static public void Main()
{
int n = 4;
Console.WriteLine(binomialCoeffSum(n));
}}
// This code is contributed by vt_m.
PHP
JavaScript
`
Output:
16