A basket has 5 bananas and 4 mangoes. In how many ways can you make a…

Each individual banana and each individual mango is treated as a distinct, identifiable item for counting selections (the standard convention in this type of selection-counting problem) — so a selection is defined by exactly which bananas and mangoes are picked, not merely how many of each type.

For a group of n distinct items, each item can independently be included or excluded, giving 2n total subsets; excluding the empty subset leaves 2n - 1 non-empty selections. When two independent groups each require at least one item to be selected, the total number of valid combined selections is the product of each group's non-empty-subset count, by the multiplication principle for independent choices.

1. Treat the 5 bananas as distinct items; each can be included or excluded, giving 25 = 32 subsets in total.

2. Exclude the empty subset (selecting zero bananas), leaving 25 - 1 = 31 ways to choose at least one banana.

3. Treat the 4 mangoes as distinct items; each can be included or excluded, giving 24 = 16 subsets in total.

4. Exclude the empty subset (selecting zero mangoes), leaving 24 - 1 = 15 ways to choose at least one mango.

5. Since the banana selection and the mango selection are made independently, multiply the two counts: 31 × 15 = 465.

Cross-check: Using inclusion-exclusion on the full set of 9 fruits: total subsets = 29 = 512. Subtract subsets with zero bananas (24 = 16) and subsets with zero mangoes (25 = 32), then add back the one subset counted twice, the empty set: 512 - 16 - 32 + 1 = 465. This matches the direct calculation, confirming 465 is correct.