PC3 · Count unordered selections using combinations
Learn to count unordered selections using combinations through clear examples and targeted practice.
Alberta Grade 12 Mathematics
Permutations, Combinations, and Binomial Theorem
Choose a group when order does not matter
A selection is different from an arrangement. If three students are chosen for a committee with identical roles, changing their order on a list does not create a new committee. That is a combination. Mathematics 30-1 asks you to recognize when order matters, calculate the appropriate count, and explain the choice. Start with the story before reaching for a formula.
What you will learn
Decide whether a situation calls for a combination.
Calculate the number of ways to choose r distinct items from n distinct items.
Explain the relationship between combinations and permutations.
Remove duplicate orders
An ordered selection of r items from n distinct items can be counted by a permutation. But each chosen group appears in r! different orders. Dividing by r! leaves each group once. This gives the combination count, written (rn) or nCr.
The conditions are 0≤r≤n, and n and r must be whole numbers. If roles are different—for example chair and secretary—order matters and this formula alone would undercount. Read the question carefully before calculating.
\binom{n}{r}=r!(n−r)!n!
Use symmetry as a check
Choosing r people to include is equivalent to choosing the n − r people to leave out, so (rn)=(n−rn). This can make arithmetic easier when r is close to n.
A committee count has no unit of probability; it is simply a number of possible groups. When there is a restriction, separate valid cases clearly or count all groups and subtract invalid ones. Do not count the same group twice.
(rn)=(n−rn)
Worked example
Choose a small study group
Six distinct students are available. How many different three-student study groups can be formed if every student has the same role?
Six students, three equal-role places
There are 6 × 5 × 4 = 120 ordered lists. Each group appears in 3! = 6 orders, leaving 20 distinct groups. The three boxes represent selection slots, not assigned roles.
Decide whether order matters
The same three students form the same group in any order, so use a combination.
Substitute in the formula
There are six candidates and three places.
\binom{6}{3}=3!3!6!=3⋅2⋅16⋅5⋅4=20
Interpret
The result counts distinct groups, not seating orders or named roles.
Answer: There are 20 different three-student groups.
Check: Choosing the three excluded students also gives (36)=20, as expected.
Common mistakes and how to avoid them
Using 6⋅5⋅4=120 for six students chosen in groups of three.
Correction: That counts each unordered group in 3!=6 orders. Divide by 6 to obtain 20.
Using a combination for named positions.
Correction: If chosen people receive different roles, assign those roles after selecting or use permutations.
Lesson summary
Use combinations for distinct items chosen without order.
The formula divides ordered arrangements by r!.
Check the story’s roles and constraints before counting.
Check your understanding
Question 1
How many two-person teams can be chosen from five distinct people?
10
20
25
Show answer and explanation
10
(25)=5!/(2!3!)=10. Switching the two people does not make a new team.
Key terms
Combination
A selection in which changing the order does not create a new outcome.
Factorial
For a positive whole number n, the product n!=n(n−1)⋯1; by convention 0!=1.
Published by DoAssignment. This reviewed lesson follows Alberta Grade 12 Mathematics (Math 30-1), expectation PC3. It is a study resource, not an official curriculum publication.
Before publication, content is checked for structure, mathematical or chemical notation, calculations, course boundaries, and readability. Errors can still occur, so corrections are welcomed.