Combinations without order (n choose k)
When order does not matter (groups, lottery), count with n choose k.
Worked example
A working group of 3 is chosen from 11 students. How many different groups are possible?
Worked solution
- 1.No order, no repetition: binomial coefficient
- 2.(11 über 3) = 11! / (3! · 8!) = 165
Typical mistake: Counting with order (n!/(n−k)!); order does not matter in a group.
Your turn
Three problems of exactly this type, each time with new numbers. Comma or point; fractions like 1/6 work too.
Task
A working group of 2 is chosen from 15 students. How many different groups 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: Counting with order (n!/(n−k)!); order does not matter in a group.
Frequently asked questions
What is the typical mistake in "Combinations without order (n choose k)"?
Counting with order (n!/(n−k)!); order does not matter in a group.
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.