From a basket that contains exactly 3 oranges, 5 apples and 7 mangoes (15 fruits in total), in how many different ways can a selection be made by taking any number of fruits, including possibly all 15, but not taking none?

Introduction / Context:
This is a combinations with identical objects problem, where the fruits of the same type are indistinguishable. We count selections based on how many of each fruit type are chosen, subject to stock limits and the condition that at least one fruit is selected. It helps practise counting with upper bounds and exclusion of the empty selection.

Given Data / Assumptions:

• There are 3 oranges, 5 apples and 7 mangoes.
• Fruits of the same type are considered identical for selection purposes.
• Any number of fruits can be taken, from 1 up to all 15.
• Selecting no fruit at all is not allowed.

Concept / Approach:
We count the number of possible triples (o, a, m), where o is the number of oranges chosen, a is the number of apples and m is the number of mangoes. Each variable is bounded between 0 and its stock limit. The total number of such triples, including the case where all three counts are zero, is simply the product of the possible choices for each type. Then we subtract the one disallowed case where no fruit is chosen.

Step-by-Step Solution:
Step 1: Let o be the number of oranges chosen. Since there are 3 oranges, o can be 0, 1, 2 or 3. This gives 4 possible values.Step 2: Let a be the number of apples chosen. Since there are 5 apples, a can be 0, 1, 2, 3, 4 or 5. This gives 6 possible values.Step 3: Let m be the number of mangoes chosen. Since there are 7 mangoes, m can be any integer from 0 to 7 inclusive, giving 8 possible values.Step 4: Without any restriction on taking at least one fruit, total selections correspond to the number of ordered triples (o, a, m) with these ranges, which is 4 * 6 * 8 = 192.Step 5: One of these triples corresponds to taking no fruit at all, namely (0, 0, 0). This selection is not allowed.Step 6: Therefore, valid selections = 192 - 1 = 191.

Verification / Alternative check:
A quick way to verify the idea is to look at a simpler example, such as 1 orange and 2 apples. In that small case, you can list all valid selections explicitly and see that the count equals (1 + 1) * (2 + 1) minus 1. Extending this logic to 3, 5 and 7 fruits per type gives the same product minus one pattern used here. No double counting occurs because each choice triple is unique.

Why Other Options Are Wrong:

• 131 and 68: These undercount and likely result from partial enumeration or forgetting that each type can have zero chosen.
• 3720: This is far too large and suggests someone treated fruits as distinct objects and used permutations, which does not match the problem statement with identical fruits per type.

Common Pitfalls:
A very common mistake is to assume fruits are distinct and use large permutation or combination counts that are not appropriate. Another error is to forget to subtract the empty selection and accept 192 instead of 191. Some students also misread the limits, such as thinking there are 15 independent choices rather than grouped by fruit type. Always focus on how many of each type you can choose.

Final Answer:
The number of different possible selections is 191.