Count Restricted Selections Without Double-Counting

Define the object before choosing a formula. Selecting an unranked group, assigning named roles and arranging a sequence are different tasks, even when they use the same people.

This original exercise counts possibilities only. It does not assume that a real selection process gives every possibility equal probability.

Start with an unranked group

Six distinct people are available: analysts A, B and C, and developers D, E and F. Select an unranked group of three containing at least one developer. How many groups qualify?

There are 6 choose 3 = 20 groups without the restriction. Exactly one group, {A, B, C}, has no developer. Subtracting it gives 20 - 1 = 19 qualifying groups.

Check by splitting the answer into disjoint cases:

The cases cannot overlap because each group has exactly one developer count. Their total is 9 + 9 + 1 = 19.

Change the recorded outcome

Each qualifying group now needs a designated presenter, chosen from its three members. If anyone may present, every group creates three distinct outcomes, so there are 19 × 3 = 57 group-and-presenter outcomes.

Now require the presenter to be a developer. Multiplying 19 by three is no longer valid: some groups contain one eligible presenter, some two and one contains three.

Weight the earlier cases by eligible presenters: 9 × 1 + 9 × 2 + 1 × 3 = 30. An independent check is to choose the developer presenter first in three ways, then select two other members from the five remaining people in ten ways. That also gives 3 × 10 = 30.

Do not divide this result by three. Each final outcome has one designated presenter; it has not been counted once for every group member. Swapping the presenter changes the outcome when the new presenter is eligible.

Explain the check, not just the answer

A concise explanation is: “I first counted membership-only groups by excluding the all-analyst group. The presenter restriction changes the object. I can count by eligible presenters within each group, or choose the presenter before the other two members.”

For another attempt, require exactly two developers and a developer presenter. The earlier table gives nine membership-only groups and two eligible presenters per group, for eighteen outcomes. Explain why this factor is constant for this new constraint before multiplying.

For the underlying mathematics, see MIT's introductory counting lecture. Keep an error log that records whether you misdefined the outcome, overlapped cases or used an invalid symmetry factor. The counting-practice comparison helps choose a free lesson, question bank or broader workbook for that specific gap.