Question
A gallery owner must choose of available paintings to hang in a single display case, where the arrangement inside the case does not matter. How many different selections are possible?
- a.
- b.
- c.
- d.
Correct answer
B.
Full reasoning
- 1The display case holds an unordered set of paintings, so order must not be counted; this calls for a combination.
- 2Compute the numerator of : ordered choices.
- 3Divide by to remove the orderings of each chosen trio: .
Why each choice is right or wrong
Choice A
This multiplies , which does not count selections at all; the count must come from a combination formula.
Choice B
Correct. unordered selections.
Choice C
This is the permutation count , which treats different hanging orders of the same three paintings as different selections.
Choice D
This computes , which allows the same painting to be chosen more than once and also counts order.
Related formula
Combination formula
Assumptions: Each item can be selected at most once and the order of selection does not matter.
Related topic and practice
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Sources
Authored for this public explainer registry. It has no database question ID and is not copied from a protected or official exam bank.
- Contemporary Mathematics, Section 7.3: Combinations — OpenStax. Accessed 2026-08-02. OpenStax textbook content is CC BY 4.0; attribute and avoid verbatim reuse beyond short cited references.
Sources and corrections
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