Given a *3 x n* board, find the number of ways to fill it with *2 x 1* dominoes.

*Example 1*

Following are all the **3** possible ways to fill up a **3 x 2** board.

*Example 2*

Here is one possible way of filling a 3 x 8 board. You have to find all the possible ways to do so.

**Examples :**

Input : 2 Output : 3 Input : 8 Output : 153 Input : 12 Output : 2131

**Defining Subproblems:**

At any point while filling the board, there are three possible states that the last column can be in:

An =No. of ways to completely fill a 3 x n board.(We need to find this)Bn =No. of ways to fill a 3 x n board with top corner in last column not filled.Cn =No. of ways to fill a 3 x n board with bottom corner in last column not filled.

**Note:** The following states are impossible to reach:

**Finding Reccurences**

**Note:** Even though **Bn** and **Cn** are different states, they will be equal for same **‘n’**. *i.e* **Bn = Cn**

Hence, we only need to calculate one of them.

**Calculating An:**

**Calculating Bn:**

**Final Recursive Relations are:**

**Base Cases:**

## C++

// C++ program to find no. of ways // to fill a 3xn board with 2x1 dominoes. #include <iostream> using namespace std; int countWays(int n) { int A[n + 1], B[n + 1]; A[0] = 1, A[1] = 0, B[0] = 0, B[1] = 1; for (int i = 2; i <= n; i++) { A[i] = A[i - 2] + 2 * B[i - 1]; B[i] = A[i - 1] + B[i - 2]; } return A[n]; } int main() { int n = 8; cout << countWays(n); return 0; }

## Java

// Java program to find no. of ways // to fill a 3xn board with 2x1 dominoes. import java.io.*; class GFG { static int countWays(int n) { int []A = new int[n+1]; int []B = new int[n+1]; A[0] = 1; A[1] = 0; B[0] = 0; B[1] = 1; for (int i = 2; i <= n; i++) { A[i] = A[i - 2] + 2 * B[i - 1]; B[i] = A[i - 1] + B[i - 2]; } return A[n]; } // Driver code public static void main (String[] args) { int n = 8; System.out.println(countWays(n)); } } // This code is contributed by anuj_67.

## Python 3

# Python 3 program to find no. of ways # to fill a 3xn board with 2x1 dominoes. def countWays(n): A = [0] * (n + 1) B = [0] * (n + 1) A[0] = 1 A[1] = 0 B[0] = 0 B[1] = 1 for i in range(2, n+1): A[i] = A[i - 2] + 2 * B[i - 1] B[i] = A[i - 1] + B[i - 2] return A[n] n = 8 print(countWays(n)) # This code is contributed by Smitha

## C#

// C# program to find no. of ways // to fill a 3xn board with 2x1 dominoes. using System; class GFG { static int countWays(int n) { int []A = new int[n+1]; int []B = new int[n+1]; A[0] = 1; A[1] = 0; B[0] = 0; B[1] = 1; for (int i = 2; i <= n; i++) { A[i] = A[i - 2] + 2 * B[i - 1]; B[i] = A[i - 1] + B[i - 2]; } return A[n]; } // Driver code public static void Main () { int n = 8; Console.WriteLine(countWays(n)); } } // This code is contributed by anuj_67.

## PHP

<?php // PHP program to find no. of ways // to fill a 3xn board with 2x1 dominoes. function countWays($n) { $A = array(); $B = array(); $A[0] = 1; $A[1] = 0; $B[0] = 0; $B[1] = 1; for ( $i = 2; $i <= $n; $i++) { $A[$i] = $A[$i - 2] + 2 * $B[$i - 1]; $B[$i] = $A[$i - 1] + $B[$i - 2]; } return $A[$n]; } // Driver Code $n = 8; echo countWays($n); // This code is contributed by anuj_67. ?>

**Output :**

153

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