Step 1: Identify what needs to be counted.
Each tank holds one particular set of 3 coral species together with one particular set of 2 sponge species. To test every possible combination, Bob needs one tank for every distinct pairing of a coral set with a sponge set. Since the order the species are placed in a tank does not matter, only which species are picked, this is a combinations problem, not a permutations problem.
Step 2: Count the ways to choose the coral species.
There are \(6\) coral species available, and \(3\) are chosen for each tank. The number of ways to do this is \[ \binom{6}{3}=\frac{6!}{3!\,3!}=\frac{6\times5\times4}{3\times2\times1}=\frac{120}{6}=20 \]
Step 3: Count the ways to choose the sponge species.
There are \(5\) sponge species available, and \(2\) are chosen for each tank. The number of ways to do this is \[ \binom{5}{2}=\frac{5!}{2!\,3!}=\frac{5\times4}{2\times1}=10 \]
Step 4: Combine the two independent choices.
Each of the \(20\) coral combinations can be paired with each of the \(10\) sponge combinations, and every such pairing is a genuinely different tank setup. By the basic counting principle, the total number of distinct tank setups is \[ 20\times10=200 \]
Step 5: Final answer.
Since every distinct coral-sponge combination needs its own tank to be tested, the minimum number of tanks required is \[ \boxed{200} \]