Unit 9: Sequences, Series, and Combinatorics · ~25 min
Objective
Apply the fundamental counting principle, permutations, and combinations to count outcomes, and use combinations to compute basic probabilities.
Lesson 9.5 used n\(C_{k}\) to count binomial coefficients without asking what it was really counting. This closing lesson answers that: n\(C_{k}\) counts the number of ways to choose k items from a group of n, and it's one of three counting tools — alongside the counting principle and permutations — that turn "how many ways can this happen" into arithmetic.
The fundamental counting principle
If one task can be done in m ways and a second, independent task can be done in n ways, the two together can be done in m × n ways. This extends to any number of stages: multiply the number of choices at each stage.
3 choices at stage one, 2 at stage two — 3 × 2 = 6 total outcomes
Permutations
A permutation counts arrangements where order matters — picking a president then a vice president from a club gives a different result than picking the same two people in the opposite order. The number of ways to arrange r items chosen from a set of n is:
Permutationsn\(P_{r}\) = \(\frac{n!}{(n-r)!}\)
Combinations
A combination counts selections where order doesn't matter — choosing a 3-person committee gives the same committee no matter what order the names were picked in. Because each group of r items could be arranged in r! different orders, dividing the permutation count by r! removes that overcounting:
This is the exact same formula as the binomial coefficient from Lesson 9.5 — n\(C_{k}\) in the Binomial Theorem is literally counting the number of ways to choose which k of the n factors of (a+b)(a+b)…(a+b) contribute a b.
A quick test: does order matter?
Situation
Order matters?
Tool
Ranking 1st, 2nd, 3rd place from 8 runners
Yes
Permutation
Choosing a 3-person committee from 8 people
No
Combination
Assigning distinct roles (chair, secretary) to 2 of 8 people
Yes
Permutation
Basic probability with combinations
When every outcome is equally likely, the probability of an event is the number of favorable outcomes divided by the total number of possible outcomes — and both counts are often found using combinations.
Probability
P(event) = \(\frac{favorable outcomes}{total outcomes}\)
Worked Example 1 · Permutation
ProblemIn how many ways can a president, vice president, and treasurer be chosen from a club of 10 members?
1Order matters (the roles are distinct), so use n\(P_{r}\) with n=10, r=3.
210\(P_{3}\) = 10!/(10−3)! = 10!/7! = 10 · 9 · 8.
720 ways
Worked Example 2 · Combination and probability
ProblemA committee of 4 is chosen at random from a group of 6 men and 5 women (11 total). Find the probability the committee is all women.
1Total ways to choose 4 from 11 (order doesn't matter): 11\(C_{4}\) = 11!/(4!7!) = 330.
2Ways to choose all 4 from the 5 women: 5\(C_{4}\) = 5!/(4!1!) = 5.
3P(all women) = 5/330 = 1/66.
P = 1/66
Guided practice
1.A restaurant offers 4 appetizers, 6 entrées, and 3 desserts. How many different 3-course meals are possible?
Show answer
Fundamental counting principle: 4 × 6 × 3 = 72.
2.How many ways can first, second, and third place ribbons be awarded in a race with 8 runners?
Show answer
Order matters: 8\(P_{3}\) = 8!/5! = 8·7·6 = 336.
3.How many different 5-person committees can be formed from a group of 12 people?
Show answer
Order doesn't matter: 12\(C_{5}\) = 12!/(5!7!) = 792.
4.A standard deck has 52 cards. In how many ways can you draw a 5-card hand containing exactly 3 aces (there are 4 aces total) and 2 non-aces (48 non-aces)?
Show answer
Choose the aces and non-aces separately, then multiply: 4\(C_{3}\) × 48\(C_{2}\) = 4 × 1128 = 4512.
5.A bag has 7 red and 3 blue marbles. Two marbles are drawn at random without replacement. Find the probability both are blue.
Show answer
Total ways to choose 2 from 10: 10\(C_{2}\) = 45. Ways to choose 2 blue from 3: 3\(C_{2}\) = 3. P = 3/45 = 1/15.