Tricky Combinations Examples

Real counting problems rarely ask for a single, plain combination. They often split a selection across several groups, force or forbid a particular member, or hide these conditions inside ordinary-sounding language. This lesson combines the formula and identities from the last two lessons into a small toolkit for those harder cases, and shows how to check the result against a smaller version of the same problem.

What you will learn

• Combine multiple selection steps into one count
• Handle combinations with grouped or restricted items
• Check a combination count against a smaller worked case

Keep paper and a pencil nearby. Try each problem before opening a hint, and mark sections reviewed as you work through them.

Multiplying independent selection steps

Some problems ask for a selection made of several separate parts, such as choosing some members from one group and other members from an unrelated group. Because the choice made within one group has no effect on the choices available in the other group, the multiplication principle applies: compute each group's combination count on its own, then multiply the counts together.

This is the same multiplication principle used for permutations, just applied to combination counts instead of ordered counts. The key check is that the groups truly do not interact — no object can belong to both pools, and no restriction links a choice in one group to a choice in the other.

Selections with required or forbidden members

If a problem requires a specific object to be part of the chosen group, that object's seat is already decided, so it should be set aside before counting. Only the remaining seats need to be filled, and they are filled from the remaining pool: a group of size $r$ from $n$ objects that must include one fixed object reduces to choosing $r-1$ objects from the other $n-1$.

If a problem instead forbids a specific object from being chosen, remove that object from the pool entirely before applying the combination formula: a group of size $r$ from $n$ objects that must exclude one fixed object reduces to choosing $r$ objects from the other $n-1$. In both cases, the trick is the same — resolve the special object first, then count the remaining, unrestricted choice.

Checking a count against a smaller case

Before trusting a combination count on a problem with large numbers, it helps to test the same method on a small, listable version of the problem. If a rule like "remove the excluded person, then choose from what remains" gives a count that matches direct listing on, say, 4 people instead of 40, then the reasoning has been confirmed and can be trusted at the larger scale.

This habit catches a very common error: miscounting how many objects remain after an object is fixed or removed. A small-case check takes only a moment and is far cheaper than working through an entire contest problem with a subtly wrong pool size.

See the reasoning unfold

Pause after each step and predict what comes next.

Common mistakes to catch

• Adding the combination counts from separate groups instead of multiplying them.
• Forgetting to remove a required or excluded person from the pool before applying the combination formula to the remaining seats.
• Skipping a small-case check that would have caught an error in how many people remain after fixing or removing one person.

Try with support

Guided practice

Choose a strategy, take a first step, and use the hints when you need direction.

Put it into practice

Independent practice

Work on your own first. If you reveal a solution, work through it and mark it reviewed.

Bring it together

• When a problem selects from separate independent groups, compute each group's combination count and multiply them.
• A required or excluded person can be set aside first, reducing the problem to choosing the remaining seats from the remaining pool.
• Checking a method on a small, listable case catches errors before trusting it on a larger problem.