The number of possible permutations for the assessment

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  • Thread starter Monoxdifly
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I think I answered the question though.In summary, there are 4 questions in the assessment and the possible questions for each question number are 13, 21, 14, and 12 respectively. To calculate the number of possible permutations for the assessment, we need to use the fundamental counting principle, which gives us 13 x 21 x 14 x 12 = 40,872 possible tests. This assumes that the question pools for each question number are non-intersecting.
  • #1
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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  • #2
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.
 
  • #3
Since that's the whole question, probably we should just assume that those groups non-intersecting. Thanks.
 

FAQ: The number of possible permutations for the assessment

What is the formula for calculating the number of possible permutations for an assessment?

The formula for calculating the number of possible permutations for an assessment is n! / (n-r)!, where n is the total number of items and r is the number of items being selected.

How do I know which number to use for n and r in the permutation formula?

The value of n represents the total number of items in the assessment, while r represents the number of items being selected. It is important to correctly identify these values in order to calculate the correct number of permutations.

What is the difference between permutations and combinations?

Permutations and combinations are both ways of counting the number of possible outcomes for a given scenario. However, permutations take into account the order of the items being selected, while combinations do not. For example, the permutations of ABC would include ABC, ACB, BAC, BCA, CAB, and CBA, while the combinations would only include ABC, ACB, and BCA.

Can the number of possible permutations for an assessment be greater than the total number of items?

No, the number of possible permutations for an assessment cannot be greater than the total number of items. This is because permutations take into account the order of the items being selected, so the number of permutations will always be equal to or less than the total number of items.

How can I use permutations to organize the order of questions on an assessment?

Permutations can be used to create a unique order for questions on an assessment. By assigning each question a number and using the permutation formula, you can generate a list of all possible orders for the questions. This can help ensure that each student receives a different version of the assessment.

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