Probability and statistics

Counting and combinatorics

Combinatorics counts possibilities without listing them. The key question is always whether order matters and whether elements can repeat.

The rules

Product rule: k steps with n₁, n₂, … options give n₁ · n₂ · … possibilities.
Arrangements of n different objects: n! = n · (n − 1) · … · 1.
Choose k of n, order matters: n! / (n − k)!.
Choose k of n, order does not matter: the binomial coefficient C(n, k) = n! / (k!(n − k)!).

Generated by the same engine as the Kollego course and solvable without a calculator. Pick an answer in your head, then open the solution.

1. How many ways can you choose 2 items from 7 (order doesn't matter)?
- a42
- b28
- c21
- d14
Show solution
Answer: 21
C(n, k) = n! / (k!·(n − k)!).
7! / (2!·5!) = 21.
2. How many ways can you choose 3 items from 6 (order doesn't matter)?
- a18
- b26
- c120
- d20
Show solution
Answer: 20
C(n, k) = n! / (k!·(n − k)!).
6! / (3!·3!) = 20.
3. How many ways can you choose 2 items from 5 (order doesn't matter)?
- a20
- b10
- c25
- d15
Show solution
Answer: 10
C(n, k) = n! / (k!·(n − k)!).
5! / (2!·3!) = 10.
4. How many ways can you choose 3 items from 4 (order doesn't matter)?
- a8
- b4
- c24
- d12
Show solution
Answer: 4
C(n, k) = n! / (k!·(n − k)!).
4! / (3!·1!) = 4.
5. How many ways can you choose 2 items from 4 (order doesn't matter)?
- a6
- b8
- c10
- d12
Show solution
Answer: 6
C(n, k) = n! / (k!·(n − k)!).
4! / (2!·2!) = 6.
6. How many ways can you choose 2 items from 6 (order doesn't matter)?
- a15
- b21
- c30
- d12
Show solution
Answer: 15
C(n, k) = n! / (k!·(n − k)!).
6! / (2!·4!) = 15.

Counting ordered arrangements when order does not matter.
Forgetting that 0! = 1.
Using the product rule when choices depend on each other in a way the task excludes.

One diagnostic, a projected score, and your first unit. No card, no time limit on the free tier.