Permutations & Combinations
The easiest way to master permutations and combinations is to stop memorizing formulas first and answer two questions: Does order matter? and Can I select the same item more than once?
Start with two questions
Question 1
Does order matter?
If changing the order creates a different outcome, think permutation.
Question 2
Can I select the same item again?
If yes, repetition/replacement is allowed. If no, the pool shrinks.
| Situation | Order matters? | Repetition / replacement? | Method |
|---|---|---|---|
| Arrange 3 people from 10 | Yes | No | Permutation without repetition |
| Create a 4-digit PIN | Yes | Yes | Permutation with repetition |
| Choose 3 people from 10 | No | No | Combination without repetition |
| Choose 3 donuts from 5 flavors; flavors may repeat | No | Yes | Combination with repetition |
The most important distinction is order
Suppose you have three letters: A, B, C and you want to select 2.
If order matters
AB and BA are different.
There are 6 outcomes.
This is a permutation.
If order does NOT matter
AB and BA represent the same selection.
There are only 3 groups.
This is a combination.
Permutation WITHOUT replacement / repetition
Suppose there are 5 people — A, B, C, D, E — and you need to choose a President, Vice-President and Secretary.
Order clearly matters because A as President and B as Vice-President is different from B as President and A as Vice-President. Also, one person cannot occupy two positions.
President: 5 choices
Vice-President: 4 choices
Secretary: 3 choices
The general formula is:
For this example: \({}_5P_3 = \dfrac{5!}{2!} = 60\)
Combination WITHOUT replacement / repetition
Now suppose the same 5 people exist, but instead of assigning positions, you simply select 3 people for a committee.
ABC is exactly the same committee as BCA or CAB. Order does not matter.
For 5 people, choose 3:
Notice the relationship:
Why divide by r!? Because permutations count every arrangement of the same group.
For example, ABC, ACB, BAC, BCA, CAB and CBA are 6 permutations, because 3! = 6, but they are all one combination: {A,B,C}.
With replacement vs without replacement
Imagine a bag containing A, B, C and D.
Without replacement
If you draw A, you do not put A back.
The next draw only has B, C and D.
With replacement
If you draw A, you put A back.
The next draw still has A, B, C and D.
So AA becomes possible.
Permutation WITH repetition / replacement
This is one of the easiest formulas.
Suppose you create a 4-digit PIN. Each position can contain 0–9, so each position has 10 choices. Repetition is allowed, so 1111 is valid.
The general formula is:
where n is the number of choices available each time and r is the number of positions.
Combination WITH repetition
This one is less intuitive.
Suppose an ice cream shop has 4 flavors — Chocolate, Vanilla, Strawberry, Mint — and you want 3 scoops. You can repeat flavors, but order does not matter.
Chocolate–Vanilla–Strawberry is the same final selection as Vanilla–Strawberry–Chocolate.
For 4 flavors and 3 scoops:
Why not n^r?
Because n^r counts order. It treats CVS, CSV, VCS, VSC, SCV and SVC as different. For a combination, those are the same final selection.
Repeated identical objects
There is another situation that often appears on exams. This is different from sampling with replacement.
Suppose you are asked: How many arrangements can be made from LEVEL?
LEVEL has 5 letters, but L appears twice and E appears twice.
If all letters were unique, there would be 5! = 120 arrangements. But swapping one L with the other L does not make a new word.
Selection with replacement
Example: create a 4-digit PIN where digits may repeat.
Objects already contain duplicates
Example: arrange the letters in MISSISSIPPI.
The table worth memorizing
| Order matters? | Repetition allowed? | Method | Formula |
|---|---|---|---|
| Yes | No | Permutation without replacement | \(\dfrac{n!}{(n-r)!}\) |
| Yes | Yes | Permutation with replacement | \(n^r\) |
| No | No | Combination without replacement | \(\dfrac{n!}{r!(n-r)!}\) |
| No | Yes | Combination with repetition | \(\binom{n+r-1}{r}\) |
Do not calculate first
Problem 1
From 12 engineers, choose 4 for a project team.
Order? No. Repeat? No.
Problem 2
From 12 runners, determine Gold, Silver and Bronze.
Order? Yes. Repeat? No.
Problem 3
Create a six-digit code using 0–9. Digits may repeat.
Order? Yes. Repeat? Yes.
Problem 4
Choose 6 donuts from 10 flavors. Multiple donuts of the same flavor are allowed.
Order? No. Repeat? Yes.
Use this every time
Question 1
Does order matter?
YES → permutation
NO → combination
Question 2
Can an item be selected more than once?
YES → repetition/replacement
NO → without replacement
\(P = \text{Position}\) \(C = \text{Choose}\) Replacement = Can I use it again?
Classify before calculating
The one memory sentence
Permutation = position matters.
Combination = only the group matters.
Replacement = can I use it again?
Almost every basic counting problem becomes much easier once you answer ORDER? and REPEAT? before reaching for a formula.
Use the same counting logic in Excel and Minitab
The software is fastest after you classify the problem correctly. First decide whether order matters and whether repetition is allowed; then use the matching function or expression.
Excel Functions
| Function | Syntax | Purpose | When to use it |
|---|---|---|---|
FACT | FACT(number) | Returns a factorial, such as 5! = 120. | Use when building a counting formula directly or checking factorial arithmetic. |
PERMUT | PERMUT(number, number_chosen) | Counts ordered selections without repetition. | Use when order matters and an item cannot be selected more than once. |
PERMUTATIONA | PERMUTATIONA(number, number_chosen) | Counts ordered selections with repetition. | Use when order matters and the same option can be used again, such as a PIN. |
COMBIN | COMBIN(number, number_chosen) | Counts unordered groups without repetition. | Use when only the selected group matters and each item can appear once. |
COMBINA | COMBINA(number, number_chosen) | Counts unordered selections with repetition. | Use when order does not matter and repeated choices are allowed. |
MULTINOMIAL | MULTINOMIAL(number1, [number2], ...) | Returns the factorial of the total divided by the product of the group factorials. | Use for arrangements of objects that already contain identical repeats, such as LEVEL. |
Excel Use Cases
=PERMUT(5,3) → 60. Roles are different, so order matters; no person repeats.
=COMBIN(5,3) → 10. Only the group matters; no person repeats.
=PERMUTATIONA(10,4) → 10,000. Four ordered positions, each with all 10 digits available.
=COMBINA(4,3) → 20. Scoop order does not matter and flavors can repeat.
=FACT(5) → 120. This is the raw count if all five objects are unique and all are arranged.
=MULTINOMIAL(2,2,1) → 30. The counts are L = 2, E = 2, V = 1.
Minitab Navigation
For these counting calculations, use Calc → Calculator. Store the answer in a constant (for example, K1) when you need one result, or in a column when the inputs vary by row.
| Counting task | Minitab Calculator expression | Why this option |
|---|---|---|
| Factorial | FACTORIAL(5) | Directly evaluates n!. |
| Permutation, no repetition | PERMUTATIONS(5,3) | Minitab's permutation function counts ordered choices without repeats. |
| Combination, no repetition | COMBINATIONS(5,3) | Minitab's combination function ignores order. |
| Ordered selection with repetition | 10^4 | Each of four positions independently has 10 choices. |
| Combination with repetition | COMBINATIONS(4+3-1,3) | Implements the stars-and-bars formula \(\binom{n+r-1}{r}\). |
| Repeated identical objects | FACTORIAL(5)/(FACTORIAL(2)*FACTORIAL(2)) | Divides out the duplicate L and E arrangements in LEVEL. |
FACTORIAL, PERMUTATIONS, and COMBINATIONS in Calc → Calculator. There is no need to search the Stat menu for a separate permutation or combination analysis.Exam Tips
ASQ CSSBB / CQE pattern
Expect the exam to hide the method inside the wording. Translate the question into two checks: order? and repeat?. Do that before touching the calculator.
Fast formula check
If a combination answer is larger than the corresponding permutation answer for the same n and r, something is wrong. Without repetition, \({}_nP_r = \binom{n}{r}r!\), so the permutation count is at least as large.
Do not confuse two types of repetition
“Replacement allowed” means you may select the same option again. “Repeated identical objects” means duplicates already exist in the set. They use different formulas.
Engineering use
Use combinations for sample/team selection when sequence has no meaning; use permutations when assigning ranked roles, inspection sequences, schedules, or position-specific settings.