A box contains 4 different black balls, 3 different red balls and 5 different blue balls.\nIn how many distinct ways can a selection of balls be made if every selection must include at least one black ball and at least one red ball?

Introduction / Context:
This problem tests understanding of combinations and the multiplication principle when different types of objects are present. The important twist is that we are counting selections (subsets), not arrangements, and we must satisfy conditions about including at least one ball of certain colours. Such questions are common in aptitude exams under the topic of permutations and combinations.

Given Data / Assumptions:

• There are 4 different black balls.
• There are 3 different red balls.
• There are 5 different blue balls.
• Each ball is distinct, even if colours are the same.
• A valid selection must contain at least 1 black ball and at least 1 red ball.
• Any number of blue balls, including zero, may be chosen.

Concept / Approach:
When several groups of distinct objects are independent, we can count choices from each group separately and multiply the results. For each colour we count how many subsets satisfy the condition on that colour. Non empty choices of black and red balls are needed, while blue balls are optional. The number of subsets of a set with n elements is 2^n, and the number of non empty subsets is 2^n - 1. We use this idea for each colour and then multiply the counts because any valid selection is formed by combining a choice of black balls, a choice of red balls and a choice of blue balls.

Step-by-Step Solution:
Number of ways to choose at least one black ball from 4 distinct black balls = 2^4 - 1 = 16 - 1 = 15. Number of ways to choose at least one red ball from 3 distinct red balls = 2^3 - 1 = 8 - 1 = 7. Number of ways to choose any subset of blue balls from 5 distinct blue balls (including choosing none) = 2^5 = 32. Because choices for each colour are independent, total valid selections = 15 * 7 * 32. Compute 15 * 7 = 105. Then 105 * 32 = 105 * (30 + 2) = 105 * 30 + 105 * 2 = 3150 + 210 = 3360.

Verification / Alternative check:
We can verify the approach logically. The only restrictions concern the presence of at least one black and one red ball. Our counting for black and red explicitly excludes empty choices, while blue balls are fully unrestricted. There is no overlap issue because each selection corresponds uniquely to a triple of subsets: one chosen from blacks, one from reds and one from blues. This one to one correspondence confirms that multiplying the three counts is valid. The final number 3360 is therefore consistent with the structure of the problem and the combinatorial rules used.

Why Other Options Are Wrong:

• 2240: This undercounts the selections. It usually comes from miscounting one colour or forgetting that blue choices include the empty subset.
• 2688: This is another incorrect product, often from taking 2^4 or 2^3 incorrectly or dropping some combinations.
• 5120: This is an overestimate that might come from allowing zero black or zero red balls, which breaks the condition.

Common Pitfalls:
Candidates often confuse selection with arrangement and try to use factorials, which would be wrong here. Another frequent mistake is forgetting that at least one ball of a colour means we must subtract the empty subset from 2^n. Some students also incorrectly force at least one blue ball, even though the question allows zero blue balls. Finally, careless arithmetic when multiplying 15, 7 and 32 can lead to an incorrect final value even when the logic is correct. Writing out intermediate products step by step helps to avoid such numerical errors.

Final Answer:
The total number of valid selections that include at least one black ball and at least one red ball is 3360.