Permutation and Combination Examples:
Counting problems trip up more QR2 students than almost any other topic — not because the formulas are hard, but because knowing which formula to use is confusing. These permutation and combination examples for BBA students use a simple test to remove the guesswork, followed by five fully worked problems.
This topic is part of the QR2 curriculum at UCP and directly supports probability calculations you’ll cover later in the same course.
The One Question That Decides Everything
Before solving any counting problem, ask: does the order of selection matter?
- If yes → use permutation.
- If no → use combination.
The Formulas
- Permutation: nPr = n! ÷ (n − r)!
- Combination: nCr = n! ÷ [r! × (n − r)!]
Notice combination has an extra r! in the denominator — that’s because it “cancels out” all the different orderings of the same group.
Example 1: Permutation — Arranging Students
A class president wants to arrange 3 students out of 8 for a presentation, in a specific speaking order.
Since order matters (who speaks first vs. last changes the outcome), use permutation:
8P3 = 8! ÷ 5! = 8 × 7 × 6 = 336 ways.
Example 2: Combination — Choosing a Committee
A society wants to choose a 3-member committee from 8 students, with no roles assigned.
Since order doesn’t matter (being “chosen” is the same regardless of order), use combination:
8C3 = 8! ÷ (3! × 5!) = (8 × 7 × 6) ÷ (3 × 2 × 1) = 56 ways.
Example 3: Permutation — Password Creation
How many 4-digit codes can be formed from digits 1–9 with no repetition?
Order clearly matters for a code, so: 9P4 = 9! ÷ 5! = 9 × 8 × 7 × 6 = 3,024 codes.
Example 4: Combination — Selecting Case Study Groups
A professor needs to select 4 students out of 12 to form a case-study group.
Group membership doesn’t depend on order: 12C4 = 12! ÷ (4! × 8!) = 495 ways.
Example 5: Mixed Problem — Both Concepts Together
A company is hiring 2 managers and 3 staff members from a pool of 5 manager applicants and 10 staff applicants.
- Managers: order doesn’t matter among the 2 selected → 5C2 = 10 ways.
- Staff: order doesn’t matter among the 3 selected → 10C3 = 120 ways.
- Total combined outcomes = 10 × 120 = 1,200 ways.
This example shows how real hiring or selection scenarios often combine multiple combination calculations using the multiplication rule.
Quick Reference Table
| Scenario | Order Matters? | Formula |
|---|---|---|
| Race finishing positions | Yes | Permutation |
| Choosing a team | No | Combination |
| PIN codes | Yes | Permutation |
| Lottery numbers drawn | No | Combination |
For the next step after counting, see our guide on probability basics for business students, and review counting principles in discrete math for the underlying rules.
Frequently Asked Questions
Q1: What’s the easiest way to tell permutation and combination apart?
Ask whether the order of the selected items matters. If yes, it’s a permutation; if no, it’s a combination.
Q2: What do these permutation and combination examples for BBA students prepare me for?
They build the foundation for probability, statistics, and later data analytics coursework where counting outcomes is essential.
Q3: What does the “!” symbol mean in these formulas?
It’s a factorial, meaning you multiply the number by every whole number below it down to 1. For example, 5! = 5 × 4 × 3 × 2 × 1 = 120.
Q4: Can combination and permutation values ever be equal?
Yes, when r = 1 or r = n, nPr and nCr can produce related results, though the formulas differ.
Q5: Why is combination smaller than permutation for the same n and r?
Because combination divides out all the repeated orderings of the same group, so it counts each unique group only once.
Keep Practicing
Try solving five new counting problems from your class handouts, and before calculating anything, write down whether it’s a permutation or combination case.
Have solved examples or class notes on this topic? Upload them to brainydock.site to help other UCP students. Explore more in our QR2 category.