There are three bags: the first contains 9 identical mangoes, the second contains 8 identical apples and the third contains 6 identical bananas. You may buy any number of fruits from each bag, from 0 up to all available, but you must buy at least one fruit in total. In how many distinct ways can you make such a purchase?

Introduction / Context:
This question asks you to count the number of integer solutions to a set of inequalities, under upper bounds. It is a typical stars and bars style problem with finite capacities, where you need to consider how many mangoes, apples and bananas you can purchase while respecting minimum and maximum limits.

Given Data / Assumptions:

• Bag 1: 9 identical mangoes available.
• Bag 2: 8 identical apples available.
• Bag 3: 6 identical bananas available.
• You can choose any number of fruits from each bag: from 0 up to the maximum available in that bag.
• You must buy at least one fruit in total (not all zeros).
• Two purchases are considered different if the counts of mangoes, apples or bananas differ.

Concept / Approach:
Let m be the number of mangoes, a the number of apples and b the number of bananas you buy. You need to count the number of integer triples (m, a, b) such that 0 ≤ m ≤ 9, 0 ≤ a ≤ 8, 0 ≤ b ≤ 6, and m + a + b ≥ 1. First count all possible triples with no restriction on the total (including the zero purchase), then subtract the single disallowed case where no fruit is purchased.

Step-by-Step Solution:
Number of choices for m (mangoes) = 10 (0 through 9).Number of choices for a (apples) = 9 (0 through 8).Number of choices for b (bananas) = 7 (0 through 6).Without the at least one fruit restriction, total possible triples (m, a, b) = 10 * 9 * 7 = 630.Among these, exactly one triple corresponds to buying no fruit at all: (m, a, b) = (0, 0, 0).We must exclude that one case because the problem requires buying at least one fruit.Therefore, the number of valid purchases = 630 - 1 = 629.

Verification / Alternative check:
You can view this as counting all combinations where the total fruits is between 1 and 23 inclusive because the maximum total is 9 + 8 + 6 = 23.However, summing case by case on total fruits is more tedious than the simple complement method above.

Why Other Options Are Wrong:
630 counts all possible triples, including the case of buying zero fruits, which is not allowed.432 and 431 do not correspond to any simple and correct counting argument under the given constraints.

Common Pitfalls:
Forgetting that buying no fruit is not allowed and therefore failing to subtract the (0, 0, 0) case.Confusing identical and distinct fruits; here identical fruits of a type reduce the problem to counting combinations of quantities.Misreading the question as buying exactly one fruit, which would produce only three options, not hundreds.

Final Answer:
The number of ways to make such a purchase is 629.