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Worked example

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Combinations

A public counting question that shows when to use a combination rather than a permutation for an unordered selection.

Question

A gallery owner must choose 33 of 99 available paintings to hang in a single display case, where the arrangement inside the case does not matter. How many different selections are possible?

  1. a.2727
  2. b.8484
  3. c.504504
  4. d.729729

Correct answer

B. 8484

Full reasoning

  1. 1The display case holds an unordered set of paintings, so order must not be counted; this calls for a combination.
  2. 2Compute the numerator of C(9,3)C(9, 3): 9×8×7=5049 \times 8 \times 7 = 504 ordered choices.
  3. 3Divide by 3!=63! = 6 to remove the orderings of each chosen trio: 504÷6=84504 \div 6 = 84.

Why each choice is right or wrong

Choice A

This multiplies 9×39 \times 3, which does not count selections at all; the count must come from a combination formula.

Choice B

Correct. C(9,3)=9×8×73!=5046=84C(9, 3) = \dfrac{9 \times 8 \times 7}{3!} = \dfrac{504}{6} = 84 unordered selections.

Choice C

This is the permutation count 9×8×79 \times 8 \times 7, which treats different hanging orders of the same three paintings as different selections.

Choice D

This computes 939^3, which allows the same painting to be chosen more than once and also counts order.

Related formula

Combination formula

C(n,r)=n!r!(nr)!C(n, r) = \frac{n!}{r!\,(n - r)!}

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.

  1. Contemporary Mathematics, Section 7.3: CombinationsOpenStax. Accessed 2026-08-02. OpenStax textbook content is CC BY 4.0; attribute and avoid verbatim reuse beyond short cited references.
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Sources and corrections

Sources last checked 2026-08-07

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