Counting selections from bounded supplies: In how many ways can fruit be selected from 3 bananas, 4 apples, and 2 oranges, where each item can be chosen 0 up to its available count?

Introduction / Context:
This is a standard “selections with upper bounds” problem where we count choices independently for each type and multiply. We are not forming permutations or combinations of identical items; we are counting how many different selection vectors (b, a, o) are possible given caps per type.

Given Data / Assumptions:

• Bananas available: 0–3 (4 choices).
• Apples available: 0–4 (5 choices).
• Oranges available: 0–2 (3 choices).
• Choosing any combination (including choosing none) is allowed.

Concept / Approach:
For each fruit, the number of possible counts equals (max + 1). By the rule of product for independent decisions, multiply the per-type possibilities together.

Step-by-Step Solution:

Verification / Alternative check:
Enumerating edges (all-max, all-min, mixed) confirms counts fall within the independent ranges, reinforcing 60.

Why Other Options Are Wrong:
39, 315, or 512 do not match the product of simple ranges (4*5*3). 512 typically appears for unconstrained binary choices over 9 items, which is not this setting.

Common Pitfalls:
Confusing “ways to select” with permutations; forgetting to include the zero-choice option per fruit.

Final Answer:
None of these (the correct count is 60).