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Step 1:: Identify all possible unique three-course selections from the list of courses.
C(9, 3) = \frac{9!}{3!(9-3)!} = \frac{9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{(3 \times 2 \times 1)(6 \times 5 \times 4 \times 3 \times 2 \times 1)} = 84
There are 9 courses in the list, and a doctoral student is required to select 3 courses. The number of ways to choose 3 courses out of 9 can be calculated using the combination formula: There are 84 possible unique three-course selections.
Step 2:: Compare the given course selections with the calculated 84 possible unique three-course selections.
The given course selections are: A. 683, 689, 606 B. 656, 683, 689 C. 685, 606, 685 D. 606, 665, 606 E. 656, 683, 685 F. 656, 685, 689 G. 656, 656 H. 656, 689, 606 I. 656, 656, 683 J. 683, 685, 689 K. 685, 689, 606 L. 656, 685, 606 M. 656, 683, 606 N. 683, 685, 606
Step 3:: Comment on the likelihood that EPR 656, EPR 683, and EPR 689 will be selected.
Since there are 84 possible unique three-course selections, and the given course selection B is one of them, the likelihood of EPR 656, EPR 683, and EPR 689 being selected is 1 out of 84, or approximately 1.19%.
Final Answer
There are 84 possible unique three-course selections, and the likelihood of EPR 656, EPR 683, and EPR 689 being selected is approximately 1.19%.
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