George divided 10 carrots and 6 radishes between the two bags such that no bag was left empty: we need to find the maximum number of ways in which he can divide the vegetables between the two bags.

Each of the 16 vegetables ( 10 carrots and 6 radishes) have got two options for the bags they can be put in, so the total number of ways to put 16 vegetables in the bags is $\(2 \times 2 \times 2 \times 2 \times \ldots \ldots \)$ \(16 times\). $\(\ldots \times 2=2^{16}\)$ which includes one way when first bag is left empty and one way when second bag is left empty which need to be subtracted.

Hence the number of ways of dividing the vegetables between the two bags such that no bag was left empty $\(=2^{16}-1-1=2^{16}-2\)$, so the answer is (E).