MHB The number of possible permutations for the assessment

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The assessment consists of four questions, each drawn from distinct pools of possible questions: 13 for question 1, 21 for question 2, 14 for question 3, and 12 for question 4. To calculate the total number of unique assessments without repeating questions, the fundamental counting principle applies, resulting in the formula 13 × 21 × 14 × 12. The initial suggestion of using 13 × 18 × 13 × 10 was incorrect, as it misrepresented the available question pools. It is important to note that the term "permutations" is not applicable here since the order of questions does not change. The final calculation yields a comprehensive count of possible assessments based on the distinct question pools.
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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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