MHB The number of possible permutations for the assessment

[Monoxdifly]

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?

 

[HallsofIvy]

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.

 

[Monoxdifly]

Since that's the whole question, probably we should just assume that those groups non-intersecting. Thanks.