Computes the set of positive even numbers with no reapted digits less than 1000 using cartesian products and unions of sets. Checks answer using less obtuse methods.

cartesianProduct.py

def cartesianProduct(a, b, repeatDigit = False):
    if repeatDigit:
        return {f'{aItem}{bItem}' for aItem in a for bItem in b}
    else: 
        return {f'{aItem}{bItem}' for aItem in a for bItem in b if all([char not in bItem for char in list(aItem)])}

odd = {'1','3','5','7','9'}
even = {'0','2','4','6','8'}
natEven = {'2','4','6','8'}

# odd X even
oe = cartesianProduct(odd, even)
#print(sorted(list(oe)))

# natEven X even
ee = cartesianProduct(natEven, even)
#print(sorted(list(ee)))

# odd X odd X even
ooe = cartesianProduct(cartesianProduct(odd, odd), even)
#print(sorted(list(ooe)))

# odd X even X even
oee = cartesianProduct(oe, even)
#print(sorted(list(oee)))

# even X odd X even
eoe = cartesianProduct(cartesianProduct(natEven, odd), even)
#print(sorted(list(eoe)))

# even X even X even
eee = cartesianProduct(ee, even)
#print(sorted(list(eee)))

evenWithNoRepeatedDigits = natEven.union(oe).union(ee).union(ooe).union(oee).union(eoe).union(eee)

nums = set()
for i in range(1, 1000):
    num = str(i)
    
    if len(num) == len(set(list(num))):
        if i % 2 == 0:
            nums.add(num)
print(f'Even numbers with no repeated digits in range [1, 999]:\n{sorted(list(nums), key=int)}')
print(f'Same set, but computed using unions oand cartesian products of smaller sets:\n{sorted(list(evenWithNoRepeatedDigits), key=int)}')

print('The two sets are the same: ', nums == evenWithNoRepeatedDigits)