diff --git a/Dynamic Programming/The Coin Change Problem/zamil.cs b/Dynamic Programming/The Coin Change Problem/zamil.cs
new file mode 100644
index 0000000..94ab9f4
--- /dev/null
+++ b/Dynamic Programming/The Coin Change Problem/zamil.cs	
@@ -0,0 +1,48 @@
+        public static long findPermutations(int n, List<long>  c)
+        {
+
+            // The 2-dimension buffer will contain answers to this question:
+            // "how much permutations is there for an amount of `i` cents, and `j`
+            // remaining coins?" eg. `buffer[10][2]` will tell us how many permutations
+            // there are when giving back 10 cents using only the first two coin types
+            // [ 1, 2 ].
+            long[][] buffer = new long[n + 1][];
+            for (var i = 0; i <= n; ++i)
+                buffer[i] = new long[c.Count + 1];
+
+            // For all the cases where we need to give back 0 cents, there's exactly
+            // 1 permutation: the empty set. Note that buffer[0][0] won't ever be
+            // needed.
+            for (var j = 1; j <= c.Count; ++j)
+                buffer[0][j] = 1;
+
+            // We process each case: 1 cent, 2 cent, etc. up to `n` cents, included.
+            for (int i = 1; i <= n; ++i)
+            {
+
+                // No more coins? No permutation is possible to attain `i` cents.
+                buffer[i][0] = 0;
+
+                // Now we consider the cases when we have J coin types available.
+                for (int j = 1; j <= c.Count; ++j)
+                {
+
+                    // First, we take into account all the known permutations possible
+                    // _without_ using the J-th coin (actually computed at the previous
+                    // loop step).
+                    var value = buffer[i][j - 1];
+
+                    // Then, we add all the permutations possible by consuming the J-th
+                    // coin itself, if we can.
+                    if (c[j - 1] <= i)
+                        value += buffer[i - c[j - 1]][j];
+
+                    // We now know the answer for this specific case.
+                    buffer[i][j] = value;
+                }
+            }
+
+            // Return the bottom-right answer, the one we were looking for in the
+            // first place.
+            return buffer[n][c.Count];
+        }