Number of ways to select one or more balls from 6 identical green balls and 4 identical blue balls

Introduction / Context:
This question checks your understanding of combinations with repeated identical objects. The balls of the same colour are identical, so the selection is determined only by how many of each colour you choose, not which specific ball you pick. The task is to count all selections that are not empty.

Given Data / Assumptions:

• There are 6 identical green balls.
• There are 4 identical blue balls.
• A selection is described only by the numbers of green and blue balls chosen.
• We must choose at least one ball in total.
• Order of selection does not matter; only the counts matter.

Concept / Approach:
We model the selection by a pair of integers (g, b) where g is the number of green balls chosen and b is the number of blue balls chosen. Valid values are 0 to the available count for each colour. We count all possible pairs except the pair where both counts are zero, because that pair corresponds to choosing no balls.

Step-by-Step Solution:
Step 1: Possible numbers of green balls g can range from 0 to 6, inclusive, so there are 7 possible values for g.Step 2: Possible numbers of blue balls b can range from 0 to 4, inclusive, so there are 5 possible values for b.Step 3: If there were no restriction, the total number of (g, b) pairs would be 7 * 5 = 35.Step 4: Among these 35 pairs, the pair (0, 0) represents choosing no green and no blue balls, that is, an empty selection.Step 5: The question asks for one or more balls, so the empty selection must be excluded.Step 6: Valid selections = total pairs minus empty pair = 35 - 1 = 34.

Verification / Alternative check:
You can also explicitly count non empty selections by considering g from 0 to 6 and for each g count allowed b values such that g and b are not both zero. Summing these counts will also give 34. This confirms the product minus one method.

Why Other Options Are Wrong:

• 42 would correspond to some incorrect counting, possibly treating balls as distinct rather than identical.
• 31 and 22 are underestimates that suggest missed combinations, perhaps by forcing at least one of each colour, which is not required.

Common Pitfalls:
A common mistake is to think in terms of distinct balls and try to use standard combinations, which is not appropriate when balls are identical in each colour. Another frequent error is forgetting to subtract the case where no balls are selected. Always remember that one or more means every non empty selection is allowed.

Final Answer:
The total number of ways to choose one or more balls is 34, so the correct answer is 34.