To fill 8 vacancies there are 15 candidates, of whom 5 belong to the ST category. If 3 of the vacancies are reserved for ST candidates and the remaining 5 vacancies are open to all candidates, in how many different ways can the selections be made?

Introduction / Context:
This question involves selection with a reservation constraint. There are 15 candidates, including 5 ST candidates, and 8 vacancies to fill. Exactly 3 vacancies are reserved for ST candidates, and the remaining 5 are open to all, which effectively means they can be filled by any of the remaining candidates, regardless of category. This structure appears frequently in competitive exams where reserved and open positions are considered within the same selection process.

Given Data / Assumptions:
- Total candidates: 15.
- ST candidates: 5.
- Non-ST candidates: 10 (since 15 - 5 = 10).
- Total vacancies: 8.
- 3 vacancies are reserved and must be filled by ST candidates.
- The remaining 5 vacancies are open to all remaining candidates (both ST and non-ST).
- We only care about which candidates are selected, not the specific seat they occupy.

Concept / Approach:
The selection can be naturally decomposed into two stages. First, we choose the 3 ST candidates to fill the reserved seats. Next, we fill the remaining 5 open seats from all remaining candidates (which include the remaining ST candidates and all non-ST candidates). Because seats themselves are not distinguished beyond the reserved versus open category, we only use combinations for each choice and multiply the results according to the rule of product.

Step-by-Step Solution:
Step 1: Select 3 ST candidates out of the 5 available ST candidates to fill the reserved vacancies.Step 2: Number of ways to do this is 5C3 = 10.Step 3: After choosing these 3 ST candidates, the remaining candidate pool consists of 15 - 3 = 12 people (2 ST and 10 non-ST).Step 4: We now must select 5 additional candidates from this pool of 12 to fill the open vacancies.Step 5: The number of ways to select 5 out of 12 candidates is 12C5.Step 6: Compute 12C5 using the formula 12! / (5! * 7!). This simplifies to (12 * 11 * 10 * 9 * 8) / (5 * 4 * 3 * 2 * 1) = 792.Step 7: Total ways to complete the selection = 5C3 * 12C5 = 10 * 792 = 7920.

Verification / Alternative check:
You can verify the logic by considering that the reserved seats force exactly 3 ST candidates to be selected initially. After that, the open seats simply require choosing any 5 from the 12 remaining candidates. No double-counting occurs because once a candidate is selected for the reserved group, they are not available for the second selection. Recomputing 12C5 independently or using known values from combination tables confirms it is 792, giving 7920 total ways when multiplied by 10.

Why Other Options Are Wrong:
- 74841 and 14874 do not arise from any natural combination product for this structure; they reflect miscalculations of combinations or incorrect interpretations of the reservation rule.
- 10213 is not a factorable product of small combination numbers like 5C3 and 12C5 and therefore does not correspond to a correct counting argument here.

Common Pitfalls:
One common error is to treat all 8 vacancies as open and then try to enforce the ST requirement afterward, risking overcounting and inclusion of selections with fewer than 3 ST candidates. Another mistake is to assume that open vacancies cannot be filled by ST candidates, which is not stated in the question. Carefully reading that open seats are available to all candidates ensures that the remaining ST candidates are still part of the pool for the open positions.

Final Answer:
The number of ways to select candidates is 7920.