unconnect
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+"""
+We shall say that an n-digit number is pandigital if it makes use of all the
+digits 1 to n exactly once; for example, the 5-digit number, 15234, is 1 through
+5 pandigital.
+
+The product 7254 is unusual, as the identity, 39 × 186 = 7254, containing
+multiplicand, multiplier, and product is 1 through 9 pandigital.
+
+Find the sum of all products whose multiplicand/multiplier/product identity can
+be written as a 1 through 9 pandigital.
+
+HINT: Some products can be obtained in more than one way so be sure to only
+include it once in your sum.
+"""
+import itertools
+
+
+def isCombinationValid(combination):
+    """
+    Checks if a combination (a tuple of 9 digits)
+    is a valid product equation.
+
+    >>> isCombinationValid(('3', '9', '1', '8', '6', '7', '2', '5', '4'))
+    True
+
+    >>> isCombinationValid(('1', '2', '3', '4', '5', '6', '7', '8', '9'))
+    False
+
+    """
+    return (
+        int(''.join(combination[0:2])) *
+        int(''.join(combination[2:5])) ==
+        int(''.join(combination[5:9]))
+    ) or (
+        int(''.join(combination[0])) *
+        int(''.join(combination[1:5])) ==
+        int(''.join(combination[5:9]))
+    )
+
+
+def solution():
+    """
+    Finds the sum of all products whose multiplicand/multiplier/product identity
+    can be written as a 1 through 9 pandigital
+
+    >>> solution()
+    45228
+    """
+
+    return sum(
+        set(
+            [
+                int(''.join(pandigital[5:9]))
+                for pandigital
+                in itertools.permutations('123456789')
+                if isCombinationValid(pandigital)
+            ]
+        )
+    )
+
+if __name__ == "__main__":
+    print(solution())