How to Solve: Combinations Basics

How to Solve: Combinations Basics

Introduction

"Imagine you’re picking a 5-player basketball team from 12 friends—how many different teams can you make? Combinations unlock the answer in seconds, and they’re on every major exam. Let’s master them now."

What You Need To Know First

Before diving into combinations, ensure you understand: 1. Factorials – The product of all positive integers up to a number (e.g., 4! = 4 × 3 × 2 × 1 = 24). 2. Basic Counting Principle – If one event has m outcomes and another has n outcomes, the total outcomes are m × n. 3. Difference Between Permutations and Combinations – Order matters in permutations (e.g., passwords), but not in combinations (e.g., teams).

Key Vocabulary

Formulas To Know

1. Combination Formula (nCr)

Formula: [ nCr = \frac{n!}{r!(n - r)!} ] Variables: - n = Total number of items. - r = Number of items being chosen. - ! = Factorial (e.g., 5! = 5 × 4 × 3 × 2 × 1).

MEMORISE THIS – It’s not always given on exams.

2. Symmetry Property (Shortcut)

Formula: [ nCr = nC(n - r) ] Why it helps: - If r is large (e.g., 10C8), calculate 10C2 instead (fewer steps). - MEMORISE THIS – Saves time on exams.

Step-by-Step Method

Follow these steps for every combination problem:

1. Identify n and r.
2. n = Total items to choose from.

3. r = Number of items being picked.

4. Check if order matters.

5. If order does not matter → Use combinations (nCr).

6. If order does matter → Use permutations (nPr).

7. Write the formula.
   [
   nCr = \frac{n!}{r!(n - r)!}
   ]

8. Plug in n and r.

9. Example: 7C3 = 7! / (3! × 4!).

10. Simplify factorials.

11. Cancel out common terms before multiplying.

12. Example: 7! / (3! × 4!) = (7 × 6 × 5 × 4!) / (3! × 4!) = (7 × 6 × 5) / 3!.

13. Calculate the remaining multiplication/division.

14. 3! = 6 → (7 × 6 × 5) / 6 = 35.

15. Verify with symmetry (optional).

16. Check if nCr = nC(n - r). Example: 7C3 = 7C4 (both = 35).

Worked Example Using Steps

Problem: How many ways can you choose 4 students from a class of 10?

1. Identify n and r.

2. n = 10 (total students), r = 4 (students chosen).

3. Check order.

4. Order doesn’t matter (Team A,B,C,D is the same as D,C,B,A).

5. Write the formula.
   [
   10C4 = \frac{10!}{4!(10 - 4)!} = \frac{10!}{4! \times 6!}
   ]

6. Simplify factorials.

7. Cancel 6! from numerator and denominator:
   [
   \frac{10 × 9 × 8 × 7 × \cancel{6!}}{4! × \cancel{6!}} = \frac{10 × 9 × 8 × 7}{4!}
   ]

8. Calculate remaining terms.

9. 4! = 24 → (10 × 9 × 8 × 7) / 24 = 5040 / 24 = 210.

10. Verify with symmetry.

11. 10C4 = 10C6 → 210 = 210 (correct).

Answer: 210 ways.

Worked Examples

Example 1 – Basic

Problem: A pizza shop offers 6 toppings. How many 2-topping pizzas can you make?

1. n = 6 (toppings), r = 2 (toppings chosen).
2. Order doesn’t matter (pepperoni + mushroom = mushroom + pepperoni).
3. Formula: 6C2 = 6! / (2! × 4!).
4. Simplify: (6 × 5 × 4!) / (2! × 4!) = (6 × 5) / 2!.
5. Calculate: 30 / 2 = 15.

What we did and why: - Used combinations because order doesn’t matter. - Simplified factorials to avoid large multiplications.

Example 2 – Medium

Problem: A club has 9 members. How many ways can you form a 3-person committee if one member must be the president?

1. Break it down:
2. The president is fixed → Only need to choose 2 more members from the remaining 8.
3. n = 8 (remaining members), r = 2.
4. Formula: 8C2 = 8! / (2! × 6!).
5. Simplify: (8 × 7 × 6!) / (2! × 6!) = (8 × 7) / 2!.
6. Calculate: 56 / 2 = 28.

What we did and why: - Recognized the fixed member reduces the problem to a smaller combination. - Avoided the mistake of using 9C3 (which would count all possible committees, not just those with the president).

Example 3 – Exam Style

Problem: A bag contains 5 red marbles and 4 blue marbles. How many ways can you pick 3 marbles with exactly 2 red and 1 blue?

1. Break into parts:
2. Choose 2 red marbles from 5 → 5C2.
3. Choose 1 blue marble from 4 → 4C1.
4. Multiply the results (Counting Principle):
5. Total ways = 5C2 × 4C1.
6. Calculate 5C2:
7. 5C2 = (5 × 4) / 2! = 10.
8. Calculate 4C1:
9. 4C1 = 4.
10. Multiply: 10 × 4 = 40.

What we did and why: - Split the problem into independent choices (red and blue marbles). - Used the Counting Principle to combine results.

Common Mistakes

Exam Traps

1-Minute Recap

"Alright, let’s lock this in. Combinations are for when order doesn’t matter—like teams, committees, or pizza toppings. The formula is nCr = n! / (r! × (n - r)!). Always: 1. Identify n (total) and r (chosen). 2. Check if order matters—if it does, you’re in permutation land. 3. Simplify factorials before multiplying. 4. Use symmetry (nCr = nC(n - r)) to save time. 5. For word problems, break them into cases (e.g., ‘exactly 2 red’ = 5C2 × 4C1).

Tonight, practice 3 problems: one basic, one with a fixed choice, and one with ‘exactly’ or ‘at least.’ You’ve got this—go ace that exam!