I'm curious what the runtime complexity would be of a naïve exhaustive search algorithm.

Here is an example:

```
def maxNumberExhaustive(self, nums1, nums2, k, current, best):
if len(current) == k:
if self.listCompare(current, best):
best[:] = current
return
elif len(current) < k:
for i in xrange(len(nums1)):
current.append(nums1[i])
self.maxNumberExhaustive(nums1[i+1::], nums2, k, current, best)
current.pop()
for i in xrange(len(nums2)):
current.append(nums2[i])
self.maxNumberExhaustive(nums1, nums2[i+1::], k, current, best)
current.pop()
```

Would it be m+n choose k, i.e. the number of subsets of length k from m + n elements?

(I know this is not an efficient solution, I'm just having trouble analyzing the brute force solution)