Welcome to Permutations and Combinations!
Hey there! Ready to dive into one of the most interesting topics in Mathematics? This chapter is all about counting. Not just 1, 2, 3... but counting the number of ways things can happen. How many ways can you arrange your favourite books? How many different teams can be formed for a project? How many possible passwords can you create?
This might sound tricky, but it's like learning the rules of a new game. Once you get the hang of it, you'll see it everywhere! We'll break it down step-by-step with simple examples. You've got this!
The Building Blocks: Counting Principles
Before we jump into the fancy stuff, we need to master two basic rules. These are the foundations for everything else in this chapter.
1. The Addition Rule (The 'OR' Rule)
Think of the Addition Rule when you have to make one choice from different groups. If you can do task A in 'm' ways OR task B in 'n' ways, and you can't do both at the same time, then the total number of ways to do one of the tasks is $$m + n$$.
Memory Aid: If you see the word 'OR' or have to choose from mutually exclusive options, you should probably ADD.
Example: A restaurant offers 5 different pasta dishes and 4 different pizzas for its main course. You can only choose ONE main course. How many choices do you have?
You can choose a pasta dish OR a pizza.
Number of choices = (Ways to choose pasta) + (Ways to choose pizza) = 5 + 4 = 9 choices.
Key Takeaway
The Addition Rule is for choosing one item from a set of separate, non-overlapping options. It's an "either/or" situation.
2. The Multiplication Rule (The 'AND' Rule)
The Multiplication Rule is for tasks that happen in stages or sequence. If you have to do task A AND then do task B, you multiply the number of ways. If task A can be done in 'm' ways and task B can be done in 'n' ways, the total number of ways to do both tasks is $$m \times n$$.
Memory Aid: If you see the word 'AND', or have a process with multiple steps, you should probably MULTIPLY.
Example: You are creating an outfit. You have 3 shirts and 4 pairs of trousers. How many different outfits can you make?
You need to choose a shirt AND a pair of trousers.
Number of outfits = (Ways to choose a shirt) x (Ways to choose trousers) = 3 x 4 = 12 outfits.
Key Takeaway
The Multiplication Rule is for processes with multiple steps that follow each other. It's an "and then" situation.
Common Mistake to Avoid
The most common mistake is mixing up these two rules! Always ask yourself: "Am I choosing one option from separate piles (OR)? Or am I making a sequence of choices, one from each pile (AND)?"
A Quick Power-Up: Factorials (!)
Before we move on, let's learn a super useful notation called factorial. It's just a shorthand way of writing a specific multiplication.
The factorial of a non-negative integer 'n', written as n!, is the product of all positive integers less than or equal to n.
$$ n! = n \times (n-1) \times (n-2) \times ... \times 3 \times 2 \times 1 $$
Examples:
$$ 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 $$
$$ 3! = 3 \times 2 \times 1 = 6 $$
$$ 1! = 1 $$
Super Important Special Case: By definition, $$0! = 1$$. This might seem weird, but it's crucial for our formulas to work correctly. Just remember it!
Permutations: When Order is Everything!
Okay, now for the first of our two big ideas. Don't worry if this seems tricky at first, it's all about the examples.
What is a Permutation?
A permutation is an ARRANGEMENT of objects in a specific, definite order. The keyword here is ORDER. If the order changes, it's a different permutation.
Real-world Analogy: A Race
Imagine a race with three runners: Alice, Bob, and Carol. The finishing order matters!
- Alice 1st, Bob 2nd, Carol 3rd is one outcome.
- Bob 1st, Alice 2nd, Carol 3rd is a completely DIFFERENT outcome.
The Permutation Formula
We want to find the number of ways to arrange 'r' objects, chosen from a set of 'n' distinct objects. We write this as $$P(n, r)$$ (or sometimes $$^nP_r$$ or $$_nP_r$$).
The formula is: $$ P(n, r) = \frac{n!}{(n-r)!} $$
- n is the total number of objects you have to choose from.
- r is the number of objects you are arranging.
Step-by-step Example:
Question: There are 8 students in a club. How many ways can a President, a Vice-President, and a Treasurer be chosen?
Step 1: Does order matter?
Yes! Being chosen as President is different from being chosen as Vice-President. So, this is a permutation problem.
Step 2: Identify 'n' and 'r'.
We are choosing from a total of 8 students, so $$n=8$$.
We are filling 3 specific positions, so $$r=3$$.
Step 3: Apply the formula.
$$ P(8, 3) = \frac{8!}{(8-3)!} = \frac{8!}{5!} $$
$$ = \frac{8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{5 \times 4 \times 3 \times 2 \times 1} $$
Notice how the 5! on the top and bottom cancel out!
$$ = 8 \times 7 \times 6 = 336 $$
So, there are 336 different ways to choose the officers.
Problem Type: Keeping Items Together
A common exam question involves arranging items where some must stay next to each other.
Example: How many ways can 4 boys and 3 girls be arranged in a row if all 3 girls must sit together?
The "Glue" Method:
Step 1: Treat the group that must be together as ONE single item. "Glue" the 3 girls together. Now, instead of 7 people, we have 4 boys + (1 group of girls) = 5 items to arrange.
Step 2: Arrange these 'glued' items. The number of ways to arrange these 5 items is $$5! = 120$$.
Step 3: Now, consider the 'glued' group. The 3 girls within their group can also arrange themselves! The number of ways to arrange the 3 girls is $$3! = 6$$.
Step 4: Use the Multiplication Rule. For each of the 120 arrangements of the main group, there are 6 arrangements within the girls' group. Total arrangements = (Ways to arrange the 5 items) x (Ways to arrange the girls inside their group) = $$5! \times 3! = 120 \times 6 = 720$$ ways.
Quick Review: Permutations
Keywords: Arrange, Order, Rank, Position, Line up
Key Idea: Order matters!
Formula: $$ P(n, r) = \frac{n!}{(n-r)!} $$
Combinations: It's All About the Group!
Now for our second big idea. This is what you use when the order of selection is irrelevant.
What is a Combination?
A combination is a SELECTION of objects where the order does not matter. It's only about which objects are chosen, not the order in which they were picked.
Real-world Analogy: Pizza Toppings
You're ordering a pizza with three toppings from a list. Does it matter if you say "pepperoni, mushrooms, and olives" versus "olives, pepperoni, and mushrooms"? No! You get the same pizza. The final group of toppings is all that matters. This is a combination.
The Combination Formula
We want to find the number of ways to select 'r' objects from a set of 'n' distinct objects. We write this as $$C(n, r)$$ (or $$^nC_r$$, $$_nC_r$$, or $$\binom{n}{r}$$).
The formula is: $$ C(n, r) = \frac{n!}{r!(n-r)!} $$
- n is the total number of objects you have to choose from.
- r is the number of objects you are choosing.
The Link Between Permutation and Combination
Look closely at the formula! It's just the permutation formula divided by $$r!$$.
$$ C(n, r) = \frac{P(n, r)}{r!} $$
Why? A permutation counts every different arrangement. To get the combination, we take all the arrangements ($$P(n,r)$$) and divide by the number of ways each small group can be arranged ($$r!$$) to remove the duplicates caused by order. This is a really important idea!
Step-by-step Example:
Question: There are 10 students. How many ways can a committee of 4 students be chosen?
Step 1: Does order matter?
No! A committee of Alice, Bob, Carol, and Dave is the exact same committee as Dave, Carol, Bob, and Alice. The order of selection doesn't create a new committee. So, this is a combination problem.
Step 2: Identify 'n' and 'r'.
We are choosing from a total of 10 students, so $$n=10$$.
We are choosing a committee of 4, so $$r=4$$.
Step 3: Apply the formula.
$$ C(10, 4) = \frac{10!}{4!(10-4)!} = \frac{10!}{4!6!} $$
$$ = \frac{10 \times 9 \times 8 \times 7 \times 6!}{ (4 \times 3 \times 2 \times 1) \times 6!} $$
Cancel the 6! from the top and bottom.
$$ = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = \frac{5040}{24} = 210 $$
So, there are 210 different committees possible.
Quick Review: Combinations
Keywords: Choose, Select, Pick, Group, Committee
Key Idea: Order does NOT matter!
Formula: $$ C(n, r) = \frac{n!}{r!(n-r)!} $$
Permutation vs. Combination: The Ultimate Showdown
This is the final hurdle: knowing when to use which one. Always ask yourself one simple question:
"Does changing the order of my selection create a new, different outcome?"
- If YES, it's a Permutation. (e.g., a phone password)
- If NO, it's a Combination. (e.g., a lottery ticket)
A Handy Cheat Sheet
Use PERMUTATION when:
- Arranging people in a line
- Awarding 1st, 2nd, 3rd place prizes
- Assigning specific roles (President, VP)
- Creating a password or code
Use COMBINATION when:
- Choosing a project team or committee
- Picking books to read
- Selecting lottery numbers
- Dealing a hand of cards
Did you know?
A "combination lock" is named incorrectly! The order you enter the numbers is very important, so it should really be called a "permutation lock"! Now you know more than the people who name the locks!
And that's it! You now have the fundamental tools to count possibilities like a pro. The key is to practice, read the question carefully, and always ask that all-important question about whether order matters. Good luck!