When Dr Ram arrives at his clinic, he finds 12 patients waiting. He can see only one patient at a time and exactly three of the patients leave before being seen. In how many different ways can this happen, considering which three leave and the order in which the remaining patients are seen?

Introduction / Context:
This problem combines combinations and permutations. We must decide which patients leave without being seen and in what order the remaining patients are seen by the doctor. Each patient is distinct, and the sequence of consultations matters. This is a classic arrangement problem that often appears in competitive exams.

Given Data / Assumptions:

There are 12 distinct patients.

Exactly 3 patients leave without being seen.

The remaining 9 patients are all seen by Dr Ram.

Dr Ram sees one patient at a time, so the order of these 9 patients matters.

Concept / Approach:
We split the task into two independent parts. First, we choose which 3 of the 12 patients leave. Second, we arrange the remaining 9 patients in an ordered sequence for consultation. The first part uses combinations, because leaving patients are not ordered. The second part uses permutations, because consultation order matters. The final answer is the product of these two counts.

Step-by-Step Solution:
Step 1: Choose 3 patients out of 12 to leave. This can be done in 12C3 ways.Step 2: Compute 12C3 = (12 * 11 * 10) / (3 * 2 * 1) = 220.Step 3: The remaining 9 patients will all be seen, and their order matters, so they can be arranged in 9! ways.Step 4: Compute 9! = 362880.Step 5: Total number of possible situations = 12C3 * 9! = 220 * 362880 = 79833600.

Verification / Alternative check:
We can think of first arranging all 12 patients and then marking 3 of them as leaving instead of being seen. However, marking positions for leaving among a 12-long sequence while preserving the order of the remaining patients leads to the same multiplication of 12C3 and 9!. This confirms our count without redoing the entire calculation.

Why Other Options Are Wrong:
479001600 equals 12! and would be correct only if all patients were seen, which is not the case here.
34879012 and 67800983 are random-looking numbers and do not correspond to any natural product of factorials or combinations in this context.

Common Pitfalls:
Many candidates forget to separate the choice of who leaves from the ordering of who is seen. Some simply compute 9! or 12! and ignore the leaving condition. Others may incorrectly compute 12P9 instead of 12C3 * 9!, which double counts many arrangements. Always break multi-step counting problems into logical stages and multiply only when stages are independent.

Final Answer:
The number of different possible ways is 79833600.