The number of possible permutations for the assessment

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The discussion focuses on calculating the number of possible assessments consisting of four questions, where each question can be selected from distinct pools: 13 questions for question 1, 21 for question 2, 14 for question 3, and 12 for question 4. The correct calculation, based on the fundamental counting principle, is 13 × 21 × 14 × 12, resulting in a total of 43,056 unique assessments. The term "permutations" is incorrectly used in the context of this discussion, as the order of questions does not change.

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Monoxdifly
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There are 4 questions in the assessment. In order not to repeat questions in an assessment, question 1 could be any of 13 questions, question 2 any of 21 questions, question 3 anyone of 14 questions and question 4 anyone of 12 questions. How do I calculate the number of possible permutations for the assessment? Is it 13 × 18 × 13 × 10?
 
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Monoxdifly said:
There are 4 questions in the assessment. In order not to repeat questions in an assessment, question 1 could be any of 13 questions, question 2 any of 21 questions, question 3 anyone of 14 questions and question 4 anyone of 12 questions. How do I calculate the number of possible permutations for the assessment? Is it 13 × 18 × 13 × 10?
We need to clarify. Are the 13 possible questions for #1 distinct from the 21 possible questions for #2? There are no questions in two different question pools? If so then, by the "fundamental counting principle", the number of tests, differing by at least one question, is 13(21)(14)(21).
I have no idea where you got "18 x 13 x 10". Also those are not "permutations" since you are not changing the order of the questions.
 
Since that's the whole question, probably we should just assume that those groups non-intersecting. Thanks.
 

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