# Calculate nCr using Pascal’s Triangle

• Difficulty Level : Medium
• Last Updated : 28 Mar, 2023

A useful application of Pascal’s triangle is the calculation of combinations. The formula to find nCr is n! / r! * (n – r)! which is also the formula for a cell of Pascal’s triangle.
Pascal’s triangle:

```Input: n = 5, r = 3
Output: 10
Explanation:
n! / r! * (n - r)! = 5! / 3! * (2!) = 120 / 12 = 10

Input: n = 7, r = 2
Output: 21
Explanation:
n! / r! * (n - r)! = 7! / 5! * (2!) = 42 / 2 = 21```

Approach: The idea is to store the Pascal’s triangle in a matrix then the value of nCr will be the value of the cell at nth row and rth column.
To create the pascal triangle use these two formula:

1. nC0 = 1, number of ways to select 0 elements from a set of n elements is 0
2. nCr = n-1Cr-1 + n-1Cr, number of ways to select r elements from a set of n elements is summation of ways to select r-1 elements from n-1 elements and ways to select r elements from n-1 elements.

The idea is to use the values of subproblems to calculate the answers for larger values. For example, to calculate nCr, use the values of n-1Cr-1 and n-1Cr. So DP can be used to preprocess all the values in the range.
Algorithm:

1. Create a matrix of size 1000 * 1000, assign the value of base cases, i.e. run a loop from 0 to 1000 and assign matrix[i][0] = 1, nC0 = 1
2. Run a nested loop from i = 1 to 1000 (outer loop) and the inner loop runs from j = 1 to i + 1.
3. For every element (i, j) assign the value of matrix[i][j] = matrix[i-1][j-1] + matrix[i-1][j], using the formula nCr = n-1Cr-1 + n-1Cr
4. After filling the matrix return the value of nCr as matrix[n][r]

Implementation:

## C++

 `// C++ implementation of the approach``#include ``using` `namespace` `std;` `// Initialize the matrix with 0``int` `l[1001][1001] = { 0 };` `void` `initialize()``{` `    ``// 0C0 = 1``    ``l[0][0] = 1;``    ``for` `(``int` `i = 1; i < 1001; i++) {``        ``// Set every nCr = 1 where r = 0``        ``l[i][0] = 1;``        ``for` `(``int` `j = 1; j < i + 1; j++) {` `            ``// Value for the current cell of Pascal's triangle``            ``l[i][j] = (l[i - 1][j - 1] + l[i - 1][j]);``        ``}``    ``}``}` `// Function to return the value of nCr``int` `nCr(``int` `n, ``int` `r)``{``    ``// Return nCr``    ``return` `l[n][r];``}` `// Driver code``int` `main()``{``    ``// Build the Pascal's triangle``    ``initialize();``    ``int` `n = 8;``    ``int` `r = 3;``    ``cout << nCr(n, r);``}` `// This code is contributed by ihritik`

## Java

 `// Java implementation of the approach` `class` `GFG {``    ``// Initialize the matrix with 0``    ``static` `int` `l[][] = ``new` `int``[``1001``][``1001``];` `    ``static` `void` `initialize()``    ``{` `        ``// 0C0 = 1``        ``l[``0``][``0``] = ``1``;``        ``for` `(``int` `i = ``1``; i < ``1001``; i++) {``            ``// Set every nCr = 1 where r = 0``            ``l[i][``0``] = ``1``;``            ``for` `(``int` `j = ``1``; j < i + ``1``; j++) {` `                ``// Value for the current cell of Pascal's triangle``                ``l[i][j] = (l[i - ``1``][j - ``1``] + l[i - ``1``][j]);``            ``}``        ``}``    ``}` `    ``// Function to return the value of nCr``    ``static` `int` `nCr(``int` `n, ``int` `r)``    ``{``        ``// Return nCr``        ``return` `l[n][r];``    ``}``    ``// Driver code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``// Build the Pascal's triangle``        ``initialize();``        ``int` `n = ``8``;``        ``int` `r = ``3``;``        ``System.out.println(nCr(n, r));``    ``}``}` `// This code is contributed by ihritik`

## Python3

 `# Python3 implementation of the approach` `# Initialize the matrix with 0``l ``=` `[[``0` `for` `i ``in` `range``(``1001``)] ``for` `j ``in` `range``(``1001``)]` `def` `initialize():``    ` `    ``# 0C0 = 1``    ``l[``0``][``0``] ``=` `1``    ``for` `i ``in` `range``(``1``, ``1001``):``        ` `        ``# Set every nCr = 1 where r = 0``        ``l[i][``0``] ``=` `1``        ``for` `j ``in` `range``(``1``, i ``+` `1``):``            ` `            ``# Value for the current cell of Pascal's triangle``            ``l[i][j] ``=` `(l[i ``-` `1``][j ``-` `1``] ``+` `l[i ``-` `1``][j])` `# Function to return the value of nCr``def` `nCr(n, r):``    ``# Return nCr``    ``return` `l[n][r]` `# Driver code``# Build the Pascal's triangle``initialize()``n ``=` `8``r ``=` `3``print``(nCr(n, r))`

## C#

 `// C# implementation of the approach` `using` `System;``class` `GFG {``    ``// Initialize the matrix with 0``    ``static` `int``[, ] l = ``new` `int``[1001, 1001];` `    ``static` `void` `initialize()``    ``{` `        ``// 0C0 = 1``        ``l[0, 0] = 1;``        ``for` `(``int` `i = 1; i < 1001; i++) {``            ``// Set every nCr = 1 where r = 0``            ``l[i, 0] = 1;``            ``for` `(``int` `j = 1; j < i + 1; j++) {` `                ``// Value for the current cell of Pascal's triangle``                ``l[i, j] = (l[i - 1, j - 1] + l[i - 1, j]);``            ``}``        ``}``    ``}` `    ``// Function to return the value of nCr``    ``static` `int` `nCr(``int` `n, ``int` `r)``    ``{``        ``// Return nCr``        ``return` `l[n, r];``    ``}``    ``// Driver code``    ``public` `static` `void` `Main()``    ``{``        ``// Build the Pascal's triangle``        ``initialize();``        ``int` `n = 8;``        ``int` `r = 3;``        ``Console.WriteLine(nCr(n, r));``    ``}``}` `// This code is contributed by ihritik`

## Javascript

 ``

Output

`56`

Complexity Analysis:

• Time Complexity: O(1).
The value of all pairs are precomputed so the time to answer the query is O(1), though some time is taken for precomputation but theoretically the precomputation takes constant time.
• Space Complexity: O(1).
Constant space is required.

### Approach:

The approach is called “Pascal’s Triangle Method”. It involves constructing Pascal’s triangle and then using the value of the corresponding cell to find nCr. The advantage of this method is that it saves time on calculating factorials by reusing previously computed values.

Steps:

1. Construct Pascal’s triangle with n+1 rows and n+1 columns.
2. Traverse to the cell corresponding to the values of n and r.
3. The value of that cell is the value of nCr.
4. Return the value of nCr.

## Python3

 `def` `pascals_triangle(n, r):``    ``triangle ``=` `[[``0``]``*``(n``+``1``) ``for` `_ ``in` `range``(n``+``1``)]``    ``for` `i ``in` `range``(n``+``1``):``        ``triangle[i][``0``] ``=` `1``    ``for` `i ``in` `range``(``1``, n``+``1``):``        ``for` `j ``in` `range``(``1``, i``+``1``):``            ``triangle[i][j] ``=` `triangle[i``-``1``][j``-``1``] ``+` `triangle[i``-``1``][j]``    ``return` `triangle[n][r]` `n ``=` `5``r ``=` `3``nCr ``=` `pascals_triangle(n, r)``print``(nCr)`

Output

`10`

Time Complexity: O(n^2)
Auxiliary Space: O(n^2)

My Personal Notes arrow_drop_up