Choose 3 of 8: how many combinations

From 8 students, how many different ways are there to choose 3 to participate in a competition?

Step-by-step

1. Total n = 8, chosen r = 3
   First distinguish the total number from the chosen number, then check whether order matters.
2. Permutations A(8, 3) = 8! / (8 − 3)! = 336
   Permutations depend on order, so we compute by dividing factorials.
3. Combinations C(8, 3) = 8! / (3!(8 − 3)!) = 56
   Combinations do not depend on order; compared to permutations, we divide by an extra r!.
4. Order does not matter ⇒ use combinations: 56
   Deciding whether order matters is the most crucial step in this type of problem.

Why it works

Choosing people to participate in a competition does not depend on order, so it is a combination rather than a permutation. The combination formula divides the permutation by an additional factorial of the number chosen; that step treats all arrangements of the same group as one selection method. Determining whether order matters is the only easy mistake in this type of problem. First ask yourself: "Does changing the order count as a new scheme?"

Common mistakes

• Calculating combination as permutation, multiplying by an extra factorial of the number chosen.
• Reversing the judgment between "order matters" and "order does not matter".
• Calculating factorial incorrectly, for example confusing the factorial of eight with the factorial of nine.

Related

n-math-counting-permutation