Ever found yourself staring at a build-your-own-burger menu, paralyzed by the sheer number of possible creations? Or maybe you’ve wondered how many different playlists you could make from your top 20 songs. This isn’t just decision fatigue; it’s a doorway into one of the most practical and fascinating areas of IB Math: Combinatorics, the art of counting.
It’s all about figuring out ‘how many ways’ something can happen. Whether you’re customizing a phone, arranging runners in a race, or forming a committee, the principles are the same. In this post, we’re going to break down the three foundational concepts you’ll need to master this topic: the Multiplication Principle, Permutations, and Combinations. Let’s get started!
Table of Contents
The Foundation: The Multiplication Rule
Everything in combinatorics starts here. It’s the simple but powerful idea that if you have a series of independent choices to make, the total number of possibilities is found by multiplying the number of options for each choice.
Rule: The Fundamental Counting Principle
If an event can occur in $m$ ways, and a second independent event can occur in $n$ ways, then the total number of ways that both events can occur is $m \times n$. This extends to any number of independent events.
Example: The Custom Gaming PC
You’re building a custom gaming PC online. You have to make a choice for each of the following components:
- CPU: 4 options
- GPU (Graphics Card): 3 options
- RAM: 4 options
- Case: 5 options
How many unique PC configurations can you build? Since each choice is independent, we just multiply the number of options together.
$4 \text{ (CPU)} \times 3 \text{ (GPU)} \times 4 \text{ (RAM)} \times 5 \text{ (Case)} = 240$
There are 240 unique configurations possible for your gaming PC. That’s a lot of choices!
When Order is King: Permutations
Now, let’s get a bit more specific. What if we are choosing a few items from a larger group, and the order in which we choose them matters? This is called a permutation. Think about a race: finishing 1st, 2nd, and 3rd is very different from finishing 3rd, 2nd, and 1st.
Example: The Science Fair Awards
At a science fair, 12 students are finalists. How many different ways can the judges award Gold (1st), Silver (2nd), and Bronze (3rd) medals?
Let’s think about it using the multiplication principle:
- For the Gold medal, there are 12 possible students.
- Once the Gold is awarded, there are 11 students left for the Silver medal.
- After that, there are 10 students remaining for the Bronze medal.
Total ways: $12 \times 11 \times 10 = 1,320$
There are 1,320 different ways to award the top three medals. This process of selecting and arranging is a permutation.
Rule: The Permutation Formula
The number of ways to arrange $r$ items from a set of $n$ distinct items is given by the formula:
$^{n}P_r = \frac{n!}{(n-r)!}$
For our science fair example ($n=12, r=3$):
$^{12}P_3 = \frac{12!}{(12-3)!} = \frac{12!}{9!}$
$= \frac{12 \times 11 \times 10 \times \cancel{9!}}{\cancel{9!}}$
$= 12 \times 11 \times 10 = 1,320$
It gives us the exact same result!
IB Exam Tip
On an exam, look for keywords like ‘arrange’, ‘order’, ‘sequence’, ‘line up’, or ‘rank’. These are massive clues that you’re dealing with a permutation. Also, don’t forget your GDC has a dedicated nPr function to calculate this quickly!
When Order is Irrelevant: Combinations
What if the order doesn’t matter? If we’re just choosing a group of items, and the arrangement within that group is irrelevant, we’re talking about a combination. Think about picking toppings for a pizza: choosing mushrooms then pepperoni is the exact same pizza as choosing pepperoni then mushrooms.
Example: The Pizza Toppings
A pizza place offers 10 different toppings. You want to choose 4 toppings for your pizza. How many different combinations of 4 toppings are possible?
If we start by thinking of this as a permutation ($^{10}P_4$), we would get 5,040. But this overcounts! Why? Because it treats {Mushroom, Pepperoni, Olives, Onions} as a different outcome from {Pepperoni, Mushroom, Onions, Olives}, even though it’s the same pizza.
We need to divide out all these duplicate arrangements. How many ways can we arrange our 4 chosen toppings? That’s $4!$ (4 factorial).
$4! = 4 \times 3 \times 2 \times 1 = 24$
So, we take our permutation result and divide by the number of duplicates:
$\frac{^{10}P_4}{4!} = \frac{5040}{24} = 210$
There are 210 unique combinations of 4 toppings.
Rule: The Combination Formula
This logic is built right into the combination formula, which finds the number of ways to choose $r$ items from a set of $n$ distinct items:
$^{n}C_r = \binom{n}{r} = \frac{n!}{r!(n-r)!}$
For our pizza problem ($n=10, r=4$):
$^{10}C_4 = \frac{10!}{4!(10-4)!} = \frac{10!}{4!6!}$
$= \frac{10 \times 9 \times 8 \times 7 \times \cancel{6!}}{(4 \times 3 \times 2 \times 1) \times \cancel{6!}}$
$= \frac{5040}{24} = 210$
Once again, the formula confirms our logic!
Permutations vs. Combinations: The Showdown
This is the most common hurdle for students. How do you know which one to use? It all boils down to one simple question. Let’s illustrate with a clear example.
Example: Club Roles vs. Movie Group
Imagine a small club with 5 members: Ana, Ben, Carla, David, and Eva.
Scenario A: We need to select a President and a Vice President.
Does order matter? Yes! ‘Ana as President, Ben as VP’ is a different outcome from ‘Ben as President, Ana as VP’. The roles make the order significant. This is a permutation.
$^{5}P_2 = \frac{5!}{(5-2)!} = \frac{5!}{3!} = 5 \times 4 = 20$ ways.
Scenario B: We need to select two members to go to a conference.
Does order matter? No! A group of ‘Ana and Ben’ is the exact same as a group of ‘Ben and Ana’. They are both just going to the conference. The group is what matters, not the order of selection. This is a combination.
$^{5}C_2 = \frac{5!}{2!(5-2)!} = \frac{5!}{2!3!} = \frac{5 \times 4}{2 \times 1} = 10$ ways.
The Ultimate Litmus Test
When faced with a problem, always ask yourself: “If I choose the items and then swap their positions, do I get a new, distinct outcome?”
- If YES (like President/VP), it’s a Permutation.
- If NO (like the conference group), it’s a Combination.
Keywords like ‘select’, ‘choose’, ‘group’, or ‘committee’ often point to combinations, but always apply the litmus test to be sure!
Key Takeaways & Your Turn
And that’s the core of counting methods! It all builds from one step to the next:
- The Multiplication Principle is your foundation for any problem involving a sequence of independent choices.
- Permutations are for counting arrangements where order is a key feature.
- Combinations are for counting groups where order is completely irrelevant.
The trickiest part is training your brain to read a question carefully and identify which tool is the right one for the job. The only way to get good at this is through practice, so let’s get to it!
Have questions or want to discuss a problem? Share your thoughts in the comments below! Engaging with the material and your peers is a fantastic way to deepen your understanding and analytical skills in mathematics.
Additional Practice
1. The Cafe Combo: A cafe offers 5 types of sandwiches, 4 types of side salads, and 6 different drinks. How many different meal combinations (one sandwich, one salad, one drink) can be made?
Show Answer
Solution: This is a direct application of the Fundamental Counting Principle. We multiply the number of options for each independent choice.
$5 \text{ (sandwiches)} \times 4 \text{ (salads)} \times 6 \text{ (drinks)} = 120$
There are 120 different meal combinations.
2. The PIN Code: How many 4-digit PIN codes can be created using the digits 0-9 if no digit can be repeated?
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Solution: The order of digits in a PIN matters, so this is a permutation. We are arranging 4 digits from a set of 10.
Using the Multiplication Principle:
$10 \text{ (choices for 1st digit)} \times 9 \text{ (for 2nd)} \times 8 \text{ (for 3rd)} \times 7 \text{ (for 4th)} = 5,040$
Using the permutation formula: $^{10}P_4 = 5,040$.
3. The Bookshelf: In how many different ways can you arrange 6 distinct books on a single shelf?
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Solution: The order of books on a shelf creates a new arrangement. This is a permutation of all 6 items. We can think of this as $6!$.
$6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720$
There are 720 ways to arrange the books.
4. The Race Medals: In a race with 10 athletes, how many ways can the Gold, Silver, and Bronze medals be awarded?
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Solution: The order of finish is crucial (Gold is different from Silver). This is a permutation of choosing and arranging 3 athletes from 10.
$^{10}P_3 = \frac{10!}{(10-3)!} = 10 \times 9 \times 8 = 720$
There are 720 ways to award the medals.
5. The Seating Arrangement: How many different ways can 5 people sit in 5 chairs arranged in a single row?
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Solution: This is an arrangement of all 5 people, so it’s a permutation of all 5 items, which is $5!$.
$5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$
There are 120 different seating arrangements.
6. The Movie Night: You want to choose 3 movies to watch from a list of 12. How many different sets of 3 movies can you choose?
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Solution: The order in which you choose the movies doesn’t matter; you’re just forming a group of 3 movies. This is a combination.
$^{12}C_3 = \frac{12!}{3!(12-3)!} = \frac{12!}{3!9!} = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = 2 \times 11 \times 10 = 220$
There are 220 different sets of 3 movies.
7. The Lottery: In a lottery, 6 numbers are drawn from a total of 49. How many different combinations of 6 numbers are possible?
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Solution: The order the numbers are drawn in a lottery doesn’t matter, only the final group of numbers. This is a classic combination problem.
$^{49}C_6 = \frac{49!}{6!(49-6)!} = \frac{49!}{6!43!} = 13,983,816$
There are nearly 14 million possible combinations!
8. The Study Group: A teacher needs to form a study group of 4 students from a class of 20. How many different groups are possible?
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Solution: A ‘group’ implies order does not matter. This is a combination.
$^{20}C_4 = \frac{20!}{4!(20-4)!} = \frac{20!}{4!16!} = \frac{20 \times 19 \times 18 \times 17}{4 \times 3 \times 2 \times 1} = 4,845$
There are 4,845 possible study groups.
9. The Company Roles: A company with 25 employees needs to fill some roles.
a) How many ways can they choose a CEO, a COO, and a CFO?
b) How many ways can they choose a 3-person advisory board?
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Solution:
a) The roles CEO, COO, and CFO are unique. The order of selection matters (Person A as CEO is different from Person A as COO). This is a permutation.
$^{25}P_3 = 25 \times 24 \times 23 = 13,800$ ways.
b) For an advisory board, all members have the same standing. The order of selection does not matter. This is a combination.
$^{25}C_3 = \frac{25 \times 24 \times 23}{3 \times 2 \times 1} = 2,300$ ways.
10. IB Exam Style Question: A committee of 4 people is to be selected from a group of 8 seniors and 5 juniors.
a) How many different committees can be formed in total?
b) How many committees can be formed if the committee must consist of exactly 2 seniors and 2 juniors?
c) How many committees can be formed if there must be at least one junior on the committee?
Show Answer
Solution:
a) For the total number of committees, we are just choosing 4 people from the total group of $8+5=13$ people. Order doesn’t matter, so it’s a combination.
Total possible committees = $^{13}C_4 = \frac{13!}{4!9!} = 715$.
b) This is a multi-step problem. We need to choose 2 seniors from 8 AND choose 2 juniors from 5. We use combinations for each choice and the multiplication principle to combine them.
Ways to choose seniors: $^{8}C_2 = \frac{8!}{2!6!} = 28$
Ways to choose juniors: $^{5}C_2 = \frac{5!}{2!3!} = 10$
Total committees with 2 seniors and 2 juniors = $28 \times 10 = 280$.
c) ‘At least one’ is a clue for using an indirect method. The opposite of ‘at least one junior’ is ‘zero juniors’ (i.e., a committee of all seniors). We can find the total committees and subtract the unwanted ones.
Total committees (from part a) = 715.
Committees with zero juniors (all 4 are seniors): Choose 4 seniors from 8. $^{8}C_4 = \frac{8!}{4!4!} = 70$.
Committees with at least one junior = Total committees – Committees with no juniors.
$715 – 70 = 645$.