Can you give us an detailed example of a score, with details on how it could vary?
Frankly I cannot understand the question.
Given a personality test (the NEO-PI-R) with 30 scales (6 sub-domains or facets per domain of the Big Five), and 3 levels of scores possible per scale (high, medium, or low), how many different arrangements of 30 scores are possible? That is, how many different combinations of 30 scores are possible? Seems that 90! is way too large a number (52 quadrillion), as is 3^30. Any help available on this? Thanks!
Yes, thank you. There are 3 possible scores or score levels: High, Average, and Low. These apply to each of the 30 subdomains or scales. There are 5 broad domains (Extraversion, Agreeableness, Conscientiousness, Neuroticism, and Openness to Experience) each having 6 subdomains or facets of the broad trait domains. For example, Extraversion's 6 subdomains are E1=Warmth, E2=Gregariousness, E3=Assertiveness, E4=Activity, E5=Excitement-Seeking, and E6=Positive Emotions). By one's response to questionnaire items designed to measure E1, one scores either high, average or low in Warmth.
Each of the other broad domains (like Agreeableness) also have 6 subdomains or scales. So after completing the entire questionnaire, a 30-trait (30-scale) profile is generated. Each facet or scale is reported as high, average or low on each of the 30 scales.
I am wanting to know how many different profile configurations are possible. The variables appear to me to be: 1) the 30 scales and 2) the 3 score levels- high, average and low. Person X may obtain a profile such as this: E1=H, E2=Av, E3=L, E4=Av, E5=H, E6=Av, A1=L, A2=Av, A3=H, A4=Av, A5=H, A6=Av, C1=H, C2=L, C3=Av, C4=H, C5=H, C6=H, N1=L, N2=L, N3=Av, N4=L, N5=Av, N6=H, O1=H, O2=H, O3=Av, O4=Av, O5=L, O6=H. Person Y may (theoretically) obtain the exact same results, but with one variation: E1=Av instead of H.
I am wanting to know the total possible number of different profiles that can be obtained. Or, to state it differently, how many different 30-score combinations are possible?
Thank you for your response. But I must tell you that it does not help me understand.
Do not be put off by that.
I have a brother who is “Chief of Psychology” at a medical school.
He tells me that I don’t understand these problems.
So maybe you can find a website that understands your vocabulary.
If you would indulge me a final attempt, I will take the matter out of the realm of psychology lingo, and pose the question in purely mathematical terms.
I have a combination (not permutation) problem. Order is not important, and repetition is not allowed. The formula n!/r!(n-r)! would normally serve just fine, but if I use 90 as n (from 3 score ranges x 30 scales), and choose 30 [as r], I get a whopping 6.7313297450658e+23. This is an inflated result because it does not account for the limiting condition of allowing the 30-set combinations to consist of only one score range (high, average, or low) per scale. And it is here that I am stuck because the above formula does not take this condition into consideration.
If you have an idea for how to solve this, I'd appreciate it. Thanks again.