Combinations

Calculate combinations where order doesn't matter. Math 30-1 Alberta Grade 12.

Lesson 7.3 of Probability Distributions in Math 30-1 — Alberta curriculum lessons.

What you'll learn in this lesson

• Introduction to Combinations. Learn when order doesn't matter. Covers the nCr formula, permutations vs. combinations, handshake and scheduling problems, the symmetry property, and solving for unknowns.
• Multi-Step Combination Problems. Combine groups using the AND rule, handle restrictions and forbidden pairs, work through at-least and at-most cases, and apply combinations to standard card-hand problems.

Inside the lesson — a free preview

Permutations vs. Combinations

The Key Distinction. Permutation: an arrangement where order matters. Picking the same people in a different order counts as a different outcome.

Combination: a selection where order does not matter. Picking the same people in a different order is still the same outcome.

Quick test: If you're choosing a committee, order doesn't matter — being picked first or last doesn't change who's on the committee. If you're choosing a President, VP, and Treasurer, order absolutely matters — those are different roles.

The Combination Formula. If n objects are selected r at a time and order does not matter:

nr = (n!)/(r!(n-r)!)

This is the permutation formula divided by r! — we divide out the r! arrangements of each group that are all counted as the same selection.

Worked example: Permutation vs. Combination Side by Side

Setup. A class of 30 students needs to select leadership.

a) Elect a President, Vice President, and Treasurer. Step 1: Does order matter?

Yes. President, VP, and Treasurer are three different roles. Picking Alex as President and Sam as VP is a different outcome from picking Sam as President and Alex as VP.

Step 2: Use permutations. Identify n and r:

n = 30 (total students to choose from), r = 3 (three different roles to fill — each role is distinct, so order matters).

_30P_3 = (30!)/((30-3)!) = (30!)/(27!) = 30 × 29 × 28 = 24,360

→ 24,360 ways

b) Elect a committee of 3. Step 1: Does order matter?

No. A committee of Alex, Sam, and Jordan is the same regardless of what order they were picked.

Step 2: Understand why we divide. From part a) we got 24,360 arrangements. But for every group of 3 people, there are 3! = 6 ways to arrange them — and all 6 count as the same committee. So we divide out the overcounting:

24,360 ÷ 6 = 4,060

Step 3: Confirm using the formula. Identify n and r:

n = 30 (total students), r = 3 (spots on the committee — same 3 people in any order is still the same committee).

nr = (n!)/(r!(n-r)!)

303 = (30!)/(3!(30-3)!) = (30!)/(3! × 27!) = 4,060

→ 4,060 committees

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