# Calculate nCr using Pascal’s Triangle

A useful application of Pascal’s triangle is the calculation of combinations. The formula to find ** ^{n}C_{r}** is

**n! / r! * (n – r)!**which is also the formula for a cell of Pascal’s triangle.

Input:n = 5, r = 3

Output:10

n! / r! * (n – r)! = 5! / 3! * (2!) = 120 / 12 = 10

Input:n = 7, r = 2

Output:21

**Approach:** The idea is to store the Pascal’s triangle in a matrix then the value of ** ^{n}C_{r}** will be the value of the cell at

**n**row and

^{th}**r**column.

^{th}Below is the implementation of the above approach:

`# 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): ` ` ` ` ` `# Build the Pascal's triangle ` ` ` `initialize() ` ` ` ` ` `# Return nCr ` ` ` `return` `l[n][r] ` ` ` `# Driver code ` `n ` `=` `8` `r ` `=` `3` `print` `(nCr(n, r)) ` |

*chevron_right*

*filter_none*

**Output:**

56

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