diff --git a/Dynamic Programming/Tiling Problem/README.md b/Dynamic Programming/Tiling Problem/README.md
new file mode 100644
index 00000000..1533e15c
--- /dev/null
+++ b/Dynamic Programming/Tiling Problem/README.md	
@@ -0,0 +1,31 @@
+# Tiling Problem (Cover a 2xN Board with 2x1 Tiles)
+
+## Problem Description
+
+The Tiling Problem involves finding the number of ways to cover a \(2 \times N\) board using \(2 \times 1\) tiles. Each tile can either be placed:
+- Vertically, which covers a \(1 \times 2\) area on the board.
+- Horizontally, which covers a \(2 \times 1\) area on the board.
+
+Given a positive integer `N`, the goal is to determine the number of ways to completely cover the \(2 \times N\) board using these tiles without gaps or overlaps.
+
+## Approach
+
+We can solve this problem using dynamic programming by defining `dp[i]` as the number of ways to tile a \(2 \times i\) board.
+
+### Steps
+
+1. **Base Cases**:
+   - For \( N = 1 \), there is only **1 way** to place a single tile vertically.
+   - For \( N = 2 \), there are **2 ways**: two vertical tiles or two horizontal tiles.
+
+2. **Recursive Formula**:
+   - For each \( i \geq 3 \):
+     \[
+     dp[i] = dp[i - 1] + dp[i - 2]
+     \]
+   - This formula is derived from two choices:
+     - Place a single vertical tile at the end (leaving a \(2 \times (i-1)\) board).
+     - Place two horizontal tiles at the end (leaving a \(2 \times (i-2)\) board).
+
+3. **Final Result**:
+   - The number of ways to tile a \(2 \times N\) board is stored in `dp[N]`.
\ No newline at end of file
diff --git a/Dynamic Programming/Tiling Problem/program.c b/Dynamic Programming/Tiling Problem/program.c
new file mode 100644
index 00000000..8585fadd
--- /dev/null
+++ b/Dynamic Programming/Tiling Problem/program.c	
@@ -0,0 +1,28 @@
+#include <stdio.h>
+
+// Function to calculate the number of ways to tile a 2xN board
+int tilingProblem(int n) {
+    if (n == 1) return 1; // Only one way to place a tile vertically
+    if (n == 2) return 2; // Two ways: two vertical tiles or two horizontal tiles
+
+    int dp[n + 1]; // Array to store the number of ways for each subproblem
+    dp[1] = 1;
+    dp[2] = 2;
+
+    for (int i = 3; i <= n; i++) {
+        dp[i] = dp[i - 1] + dp[i - 2];
+    }
+
+    return dp[n];
+}
+
+int main() {
+    int n;
+    printf("Enter the length of the board (N): ");
+    scanf("%d", &n);
+
+    int ways = tilingProblem(n);
+    printf("Number of ways to tile a 2x%d board is: %d\n", n, ways);
+
+    return 0;
+}