Combinations

Formula

The number of ways to choose $r$ objects from $n$ distinct objects without order is

The conditions are $0\le r\le n$. Choosing no objects or all objects each has one outcome.

Why divide by r factorial

The permutation count ${}_nP_r$ counts every selected group in all $r!$ possible internal orders. A combination treats those orders as the same group, so

This relationship explains the denominator rather than making it a mysterious formula.

Worked example

The same four students listed in another order remain the same committee.

Combination or permutation?

Ask whether swapping positions creates a different outcome. Choosing three pizza toppings is a combination; assigning gold, silver, and bronze among three people is a permutation.

Named roles can turn a selection problem into a two-stage process: first choose members, then assign roles.

Common mistakes

Using combinations for assigned roles. Roles create order.

Using permutations and forgetting to divide by internal orders. The same group gets overcounted.

Multiplying cases that are alternatives. Disjoint “or” cases are added.

Using complement counts without defining the universe. Start from the correct total set.

Mixing ordered favourable outcomes with unordered total outcomes. Count consistently.

Quick self-check

• Does rearranging the same selected members change the outcome?
• What are $n$ and $r$ after restrictions are applied?
• Are required or forbidden members handled before the formula?
• Should category counts multiply or case counts add?
• Would a complement be shorter?
• Is the probability sample space counted with the same order convention?