Binomial Coefficient
Last Updated :
17 Feb, 2025
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Given an integer values n and k, the task is to find the value of Binomial Coefficient C(n, k).
- A binomial coefficient C(n, k) can be defined as the coefficient of x^k in the expansion of (1 + x)^n.
- A binomial coefficient C(n, k) also gives the number of ways, disregarding order, that k objects can be chosen from among n objects more formally, the number of k-element subsets (or k-combinations) of a n-element set.
Examples
Input: n = 4, k = 2
Output: 6
Explanation: The value of 4C2 is (4 × 3) / (2 × 1) = 6.Input: n = 5, k = 2
Output: 10
Explanation: The value of 5C2 is (5 × 4) / (2 × 1) = 10.Input: n = 6, k = 3
Output: 20
Explanation: The value of 6C3 is (6 × 5 × 4) / (3 × 2 × 1) = 20.
Table of Content
Using recursion – O(2 ^ n) Time and O(n) Space
The idea is to use recursion to find C(n, k).The value of C(n, k) can be recursively calculated using the following standard formula for Binomial Coefficients. C(n, k) = C(n-1, k-1) + C(n-1, k). C(n, 0) = C(n, n) = 1. So we just need to make recursive calls of C(n-1, k-1) and C(n – 1, k). The base conditions will be when k = 0 or value of k and n be equal.
// C++ implementation to find
// Binomial Coefficient using recursion
#include <bits/stdc++.h>
using namespace std;
// Returns value of Binomial Coefficient C(n, k)
int binomialCoeff(int n, int k) {
// k can not be grater then k so we return 0 here
if (k > n)
return 0;
// base condition when k and n are equal or k = 0
if (k == 0 || k == n)
return 1;
// Recurvie add the value
return binomialCoeff(n - 1, k - 1)
+ binomialCoeff(n - 1, k);
}
int main() {
int n = 5, k = 2;
cout << binomialCoeff(n, k);
return 0;
}
// C implementation to find
// Binomial Coefficient using recursion
#include <stdio.h>
// Returns value of Binomial Coefficient C(n, k)
int binomialCoeff(int n, int k) {
// k can not be grater then k so we return 0 here
if (k > n)
return 0;
// base condition when k and n are equal or k = 0
if (k == 0 || k == n)
return 1;
// Recursive add the value
return binomialCoeff(n - 1, k - 1)
+ binomialCoeff(n - 1, k);
}
int main() {
int n = 5, k = 2;
printf("%d", binomialCoeff(n, k));
return 0;
}
// Java implementation to find
// Binomial Coefficient using recursion
class GfG {
// Returns value of Binomial Coefficient C(n, k)
static int binomialCoeff(int n, int k) {
// k can not be grater then k so we
// return 0 here
if (k > n)
return 0;
// base condition when k and n are
// equal or k = 0
if (k == 0 || k == n)
return 1;
// Recursive add the value
return binomialCoeff(n - 1, k - 1)
+ binomialCoeff(n - 1, k);
}
public static void main(String[] args) {
int n = 5, k = 2;
System.out.println(binomialCoeff(n, k));
}
}
# Python implementation to find
# Binomial Coefficient using recursion
# Returns value of Binomial Coefficient C(n, k)
def binomialCoeff(n, k):
# k can not be grater then k so we
# return 0 here
if k > n:
return 0
# base condition when k and n are equal
# or k = 0
if k == 0 or k == n:
return 1
# Recursive add the value
return binomialCoeff(n - 1, k - 1) + binomialCoeff(n - 1, k)
n = 5
k = 2
print(binomialCoeff(n, k))
// C# implementation to find
// Binomial Coefficient using recursion
using System;
class GfG {
// Returns value of Binomial Coefficient C(n, k)
static int BinomialCoeff(int n, int k) {
// k can not be grater then k so we
// return 0 here
if (k > n)
return 0;
// base condition when k and n are
// equal or k = 0
if (k == 0 || k == n)
return 1;
// Recursive add the value
return BinomialCoeff(n - 1, k - 1)
+ BinomialCoeff(n - 1, k);
}
static void Main(string[] args) {
int n = 5, k = 2;
Console.WriteLine(BinomialCoeff(n, k));
}
}
// Javascript implementation to find
// Binomial Coefficient using recursion
// Returns value of Binomial Coefficient C(n, k)
function binomialCoeff(n, k) {
// k can not be grater then k so we
// return 0 here
if (k > n)
return 0;
// base condition when k and n are equal
// or k = 0
if (k === 0 || k === n)
return 1;
// Recursive add the value
return binomialCoeff(n - 1, k - 1) + binomialCoeff(n - 1, k);
}
let n = 5, k = 2;
console.log(binomialCoeff(n, k));
Output
10
Top-Down DP (Memoization) – O(n * k) Time and O(n * k) Space
It should be noted that the above function computes the same subproblems again and again. And have two properties of Dynamic Programming:
1. Optimal Substructure:
The value of C(n, k)depends on the optimal solutions of the subproblemsC(n-1, k-1) and C(n-1, k). By adding these optimal substrutures, we can efficiently calculate the total value of C(n, k).
2. Overlapping Subproblems:
While applying a recursive approach in this problem, we notice that certain subproblems are computed multiple times. Recursion tree for n = 5 and k = 2. The function C(3, 1) is called two times. For large values of n, there will be many common subproblems.
The Binomial Coefficient C(n, k) is computed recursively, but to avoid redundant calculations, dynamic programming with memoization is used. A 2D table stores previously computed values, allowing efficient lookups instead of recalculating. If a value is already computed, it is returned directly; otherwise, it is computed recursively and stored for future use.
// C++ implementation to find
// Binomial Coefficient using memoization
#include <bits/stdc++.h>
using namespace std;
// Returns value of Binomial Coefficient C(n, k)
int getnCk(int n, int k, vector<vector<int>> &memo) {
// k can not be grater then k so we
// return 0 here
if (k > n)
return 0;
// base condition when k and n are
// equal or k = 0
if (k == 0 || k == n)
return 1;
// Check if pair n and k is already
// calculated then return it from here
if(memo[n][k] != -1) return memo[n][k];
// Recurvie add the value and store to memorize table
return memo[n][k] = getnCk(n - 1, k - 1, memo)
+ getnCk(n - 1, k, memo);
}
int binomialCoeff(int n, int k) {
// Create table for memorization
vector<vector<int>> memo(n + 1, vector<int> (k + 1, -1));
return getnCk(n, k, memo);
}
int main() {
int n = 5, k = 2;
cout << binomialCoeff(n, k);
return 0;
}
// Java implementation to find
// Binomial Coefficient using memoization
import java.util.Arrays;
class GfG {
// Returns value of Binomial Coefficient C(n, k)
static int getnCk(int n, int k, int[][] memo) {
// k cannot be greater than n so we return 0 here
if (k > n)
return 0;
// base condition when k and n are equal or k = 0
if (k == 0 || k == n)
return 1;
// Check if pair n and k is already
// calculated then return it from here
if (memo[n][k] != -1) return memo[n][k];
// Recursive add the value and store to memo table
return memo[n][k] = getnCk(n - 1, k - 1, memo)
+ getnCk(n - 1, k, memo);
}
static int binomialCoeff(int n, int k) {
// Create table for memoization
int[][] memo = new int[n + 1][k + 1];
for (int[] row : memo)
Arrays.fill(row, -1);
return getnCk(n, k, memo);
}
public static void main(String[] args) {
int n = 5, k = 2;
System.out.println(binomialCoeff(n, k));
}
}
# Python implementation to find
# Binomial Coefficient using memoization
def getnCk(n, k, memo):
# k cannot be greater than n so we return 0 here
if k > n:
return 0
# base condition when k and n are equal or k = 0
if k == 0 or k == n:
return 1
# Check if pair n and k is already
# calculated then return it from here
if memo[n][k] != -1:
return memo[n][k]
# Recursive add the value and store to memo table
memo[n][k] = getnCk(n - 1, k - 1, memo) + \
getnCk(n - 1, k, memo)
return memo[n][k]
def binomialCoeff(n, k):
# Create table for memoization
memo = [[-1 for _ in range(k + 1)] for _ in range(n + 1)]
return getnCk(n, k, memo)
n, k = 5, 2
print(binomialCoeff(n, k))
// C# implementation to find
// Binomial Coefficient using memoization
using System;
class GfG {
// Returns value of Binomial
// Coefficient C(n, k)
static int GetnCk(int n, int k, int[,] memo) {
// k cannot be greater than n so we
// return 0 here
if (k > n)
return 0;
// base condition when k and n are
// equal or k = 0
if (k == 0 || k == n)
return 1;
// Check if pair n and k is already
// calculated then return it from here
if (memo[n, k] != -1) return memo[n, k];
// Recursive add the value and store to memo table
return memo[n, k] = GetnCk(n - 1, k - 1, memo)
+ GetnCk(n - 1, k, memo);
}
static int BinomialCoeff(int n, int k) {
// Create table for memoization
int[,] memo = new int[n + 1, k + 1];
for (int i = 0; i <= n; i++)
for (int j = 0; j <= k; j++)
memo[i, j] = -1;
return GetnCk(n, k, memo);
}
static void Main() {
int n = 5, k = 2;
Console.WriteLine(BinomialCoeff(n, k));
}
}
// Javascript implementation to find
// Binomial Coefficient using memoization
function getnCk(n, k, memo) {
// k cannot be greater than n so we
// return 0 here
if (k > n)
return 0;
// base condition when k and n are
// equal or k = 0
if (k === 0 || k === n)
return 1;
// Check if pair n and k is already
// calculated then return it from here
if (memo[n][k] !== -1)
return memo[n][k];
// Recursive add the value and store to memo table
return memo[n][k] = getnCk(n - 1, k - 1, memo)
+ getnCk(n - 1, k, memo);
}
function binomialCoeff(n, k) {
// Create table for memoization
const memo = Array.from({length : n + 1},
() => Array(k + 1).fill(-1));
return getnCk(n, k, memo);
}
const n = 5, k = 2;
console.log(binomialCoeff(n, k));
Output
10
Using Bottom-Up DP (Tabulation) – O(n * k) Time and O(n * k) Space
The approach is similar to the previous one. just instead of breaking down the problem recursively, we iteratively build up the solution by calculating in bottom-up manner. Maintain a dp[][] table such that dp[i][j] stores the count all unique possible paths to reach the cell (i, j).
Base Case:
- For i = j and 0 <= i <= n , dp[i][j] = 1
- for j = 0 and 0 <= j <= min(i, k) , dp[i][j] = 1
Recursive Case:
- For i > 1 and j > 1 , dp[i][j] = dp[i-1][j-1] + dp[i-1][j]
// C++ implementation to find
// Binomial Coefficient using tabulation
#include <bits/stdc++.h>
using namespace std;
// Returns value of Binomial Coefficient C(n, k)
int binomialCoeff(int n, int k) {
vector<vector<int>> dp(n + 1, vector<int> (k + 1));
// Calculate value of Binomial Coefficient
// in bottom up manner
for (int i = 0; i <= n; i++) {
for (int j = 0; j <= min(i, k); j++) {
// Base Cases
if (j == 0 || j == i)
dp[i][j] = 1;
// Calculate value using previously
// stored values
else
dp[i][j] = dp[i - 1][j - 1] + dp[i - 1][j];
}
}
return dp[n][k];
}
int main() {
int n = 5, k = 2;
cout << binomialCoeff(n, k);
}
// C implementation to find
// Binomial Coefficient using tabulation
#include <stdio.h>
// Returns value of Binomial Coefficient C(n, k)
int binomialCoeff(int n, int k) {
int dp[n + 1][k + 1];
// Calculate value of Binomial Coefficient
// in bottom up manner
for (int i = 0; i <= n; i++) {
for (int j = 0; j <= (i < k ? i : k); j++) {
if (j == 0 || j == i)
dp[i][j] = 1;
// Calculate value using previously
// stored values
else
dp[i][j] = dp[i - 1][j - 1] + dp[i - 1][j];
}
}
return dp[n][k];
}
int main() {
int n = 5, k = 2;
printf("%d", binomialCoeff(n, k));
return 0;
}
// Java implementation to find
// Binomial Coefficient using tabulation
class GfG {
// Returns value of Binomial Coefficient C(n, k)
static int binomialCoeff(int n, int k) {
int[][] dp = new int[n + 1][k + 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, k); j++) {
if (j == 0 || j == i)
dp[i][j] = 1;
// Calculate value using previously
// stored values
else
dp[i][j]
= dp[i - 1][j - 1] + dp[i - 1][j];
}
}
return dp[n][k];
}
public static void main(String[] args) {
int n = 5, k = 2;
System.out.println(binomialCoeff(n, k));
}
}
# Python implementation to find
# Binomial Coefficient using tabulation
# Returns value of Binomial Coefficient C(n, k)
def binomialCoeff(n, k):
dp = [[0 for _ in range(k + 1)] for _ in range(n + 1)]
# Calculate value of Binomial Coefficient
# in bottom up manner
for i in range(n + 1):
for j in range(min(i, k) + 1):
# Base Cases
if j == 0 or j == i:
dp[i][j] = 1
# Calculate value using previously
# stored values
else:
dp[i][j] = dp[i - 1][j - 1] + dp[i - 1][j]
return dp[n][k]
n = 5
k = 2
print(binomialCoeff(n, k))
// C3 implementation to find
// Binomial Coefficient using tabulation
using System;
class GfG {
// Returns value of Binomial Coefficient C(n, k)
public static int BinomialCoeff(int n, int k) {
int[, ] dp = new int[n + 1, k + 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, k); j++) {
// Base Cases
if (j == 0 || j == i)
dp[i, j] = 1;
// Calculate value using previously
// stored values
else
dp[i, j]
= dp[i - 1, j - 1] + dp[i - 1, j];
}
}
return dp[n, k];
}
static void Main(string[] args) {
int n = 5, k = 2;
Console.WriteLine(BinomialCoeff(n, k));
}
}
// Javascript implementation to find
// Binomial Coefficient using tabulation
// Returns value of Binomial Coefficient C(n, k)
function binomialCoeff(n, k) {
let dp = Array.from({ length: n + 1 }, () => Array(k + 1).fill(0));
// Calculate value of Binomial Coefficient
// in bottom up manner
for (let i = 0; i <= n; i++) {
for (let j = 0; j <= Math.min(i, k); j++) {
// Base Cases
if (j === 0 || j === i) {
dp[i][j] = 1;
}
// Calculate value using previously
// stored values
else {
dp[i][j] = dp[i - 1][j - 1] + dp[i - 1][j];
}
}
}
return dp[n][k];
}
let n = 5, k = 2;
console.log(binomialCoeff(n, k));
Output
10
Using Space Optimized DP – O(n * k) Time and O(k) Space
In the previous approach using dynamic programming, we derived a relation between states as follows:
- dp[i][j] = dp[i-1][j-1]+dp[i-1][j]
We do not need to maitain whole matrix for this. We can just maintain one array of length k and add dp[j-1] every time to dp[j];
- Use a 1D array dp[] of size k+1 to store binomial coefficients, reducing space complexity to O(k).
- Set dp[0] = 1, representing nC0=1.
- Update dp[j] in reverse order, using the previous values from the same array.
- Each entry dp[j] is updated as dp[j] + dp[j-1] for each row.
- The final value of dp(n,k) is stored in dp[k], and returned.
// C++ program for space optimized Dynamic Programming
// Solution of Binomial Coefficient
#include <bits/stdc++.h>
using namespace std;
int binomialCoeff(int n, int k) {
vector<int> dp(k + 1);
// nC0 is 1
dp[0] = 1;
for (int i = 1; i <= n; i++) {
// Compute next row of pascal triangle using
// the previous row
for (int j = min(i, k); j > 0; j--)
dp[j] = dp[j] + dp[j - 1];
}
return dp[k];
}
int main() {
int n = 5, k = 2;
cout << binomialCoeff(n, k);
return 0;
}
// Java program for space optimized Dynamic Programming
// Solution of Binomial Coefficient
class GfG {
// Returns value of Binomial Coefficient C(n, k)
static int binomialCoeff(int n, int k) {
int[] dp = new int[k + 1];
// nC0 is 1
dp[0] = 1;
for (int i = 1; i <= n; i++) {
// Compute next row of pascal triangle using
// the previous row
for (int j = Math.min(i, k); j > 0; j--)
dp[j] = dp[j] + dp[j - 1];
}
return dp[k];
}
public static void main(String[] args) {
int n = 5, k = 2;
System.out.println(binomialCoeff(n, k));
}
}
# Python program for space optimized Dynamic Programming
# Solution of Binomial Coefficient
def binomialCoeff(n, k):
dp = [0] * (k + 1)
# nC0 is 1
dp[0] = 1
for i in range(1, n + 1):
# Compute next row of pascal triangle using
# the previous row
for j in range(min(i, k), 0, -1):
dp[j] = dp[j] + dp[j - 1]
return dp[k]
n = 5
k = 2
print(binomialCoeff(n, k))
// C# program for space optimized Dynamic Programming
// Solution of Binomial Coefficient
using System;
class GfG {
// Returns value of Binomial Coefficient C(n, k)
static int BinomialCoeff(int n, int k) {
int[] dp = new int[k + 1];
// nC0 is 1
dp[0] = 1;
for (int i = 1; i <= n; i++) {
// Compute next row of pascal triangle using
// the previous row
for (int j = Math.Min(i, k); j > 0; j--)
dp[j] = dp[j] + dp[j - 1];
}
return dp[k];
}
static void Main(string[] args) {
int n = 5, k = 2;
Console.WriteLine(BinomialCoeff(n, k));
}
}
// JavaScript program for space optimized Dynamic
// Programming Solution of Binomial Coefficient
function binomialCoeff(n, k) {
let dp = Array(k + 1).fill(0);
// nC0 is 1
dp[0] = 1;
for (let i = 1; i <= n; i++) {
// Compute next row of pascal triangle using
// the previous row
for (let j = Math.min(i, k); j > 0; j--) {
dp[j] = dp[j] + dp[j - 1];
}
}
return dp[k];
}
let n = 5, k = 2;
console.log(binomialCoeff(n, k));
Output
10
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