Permutations with order
When order matters (1st, 2nd, 3rd place), multiply the shrinking number of choices.
Worked example
In a contest with 8 teams, places 1 to 2 are awarded. How many different results are possible?
Worked solution
- 1.With order, no repetition: n! / (n − k)!
- 2.8! / 6! = 8 · 7 = 56
Typical mistake: Using the binomial coefficient; order matters for places.
Your turn
Three problems of exactly this type, each time with new numbers. Comma or point; fractions like 1/6 work too.
Task
In a contest with 8 teams, places 1 to 3 are awarded. How many different results are possible?
Number
Comma or point; fractions like 1/6 work too.
Formulas for this topic
- Classical: P(A) = |A| / |Ω|
- No order, no repetition: C(n, k) = n! / (k! · (n − k)!)
- With order, no repetition: n! / (n − k)!
Typical mistake: Using the binomial coefficient; order matters for places.
Frequently asked questions
What is the typical mistake in "Permutations with order"?
Using the binomial coefficient; order matters for places.
How do I know if order matters?
If a different order is a different result (places, PINs) it matters; for groups or lottery numbers it does not.