Final answer:
The student is given a problem of finding combinations with restrictions in a Middle School mathematics context. After accounting for the minimum type requirements, the collector can create 55 different combinations of the three types of monsters within the 20 containment devices.
Step-by-step explanation:
The question asks for the number of different combinations of small monster types a collector can capture using 20 containment devices, given certain restrictions on the minimum number of each type of monster they wish to capture.
Firstly, we need to account for the minimum requirements: the collector needs at least 3 grass types, 3 plant types, and 5 earth types. This equates to 3 + 3 + 5 = 11 monsters, leaving us 20 - 11 = 9 more monsters to allocate freely across the three types.
The problem can now be reframed as finding the number of non-negative integer solutions to the equation g + p + e = 9, where g, p, e represent the additional number of grass, plant, and earth-type monsters, respectively, above the minimum already taken. The formula for such a problem is given by (n + k - 1) choose (k - 1), where n is the number of items to choose from (3 types in this case) and k is the number of items to be chosen (9 additional monsters).
Plugging into the formula, we have (9 + 3 - 1) choose (3 - 1) which simplifies to 11 choose 2. Using the combination formula C(n, k) = n! / [k!(n-k)!], we find that the collector can create 55 different combinations of the small monster types for containment.