Permutations and combinations: arrangements, selections
Permutations and Combinations: Arrangements & Selections 📚
1️⃣ Arrangements (Permutations) 🔄
When the order matters, we talk about permutations. Think of arranging books on a shelf or lining up for a photo.
- Formula for arranging all n items:
$P(n)=n!$ - Formula for arranging r out of n items (order matters):
$P(n,r)=\dfrac{n!}{(n-r)!}$ - Example: 3 different coloured balls (Red, Blue, Green) in a line:
$P(3)=3!=6$ ways: RBG, RGB, BRG, BGR, GRB, GBR. - With repeated items: If 2 balls are the same colour, divide by the factorial of the repeats:
$\dfrac{n!}{k_1!k_2!\dots}$
2️⃣ Selections (Combinations) 📋
When the order does not matter, we use combinations. Imagine picking a team of 4 from 10 students.
- Formula for choosing r out of n (no order):
$C(n,r)=\binom{n}{r}=\dfrac{n!}{r!(n-r)!}$ - Example: Choose 2 out of 5 friends:
$\binom{5}{2}=10$ possible pairs. - With identical items (stars and bars): For selecting r items from n types with unlimited supply:
$\binom{n+r-1}{r}$ - Practice Tip: If you can count the arrangements and then divide by the number of ways to reorder the chosen items, you get the combinations.
3️⃣ Quick Reference Table 📊
| Scenario | Formula | Example |
|---|---|---|
| All items, order matters | $P(n)=n!$ | Arrange 4 books: $4!=24$ ways |
| Choose r, order matters | $P(n,r)=\dfrac{n!}{(n-r)!}$ | Pick 2 out of 5 students: $\dfrac{5!}{3!}=20$ ways |
| Choose r, order irrelevant | $C(n,r)=\binom{n}{r}$ | Select 3 out of 7: $\binom{7}{3}=35$ ways |
| With repeats allowed (stars & bars) | $\binom{n+r-1}{r}$ | Choose 3 candies from 4 types: $\binom{4+3-1}{3}=20$ ways |
4️⃣ Quick Practice Problems 🧩
- How many ways can 5 different shirts be worn in a row if you only wear 3?
Answer: $P(5,3)=\dfrac{5!}{2!}=60$ - From 12 students, how many 4‑person teams can be formed?
Answer: $\binom{12}{4}=495$ - Choose 2 sweets from 3 types (apple, banana, cherry) with unlimited supply. How many combinations?
Answer: $\binom{3+2-1}{2}=6$ - Arrange the letters of the word "LEVEL". How many distinct arrangements?
Answer: $\dfrac{5!}{2!}=60$ (since L appears twice)
5️⃣ Takeaway Tips ✨
- Use factorials ($n!$) as the building block.
- When order doesn't matter, divide by the factorial of the number of items chosen to remove duplicate arrangements.
- For selections with unlimited supply, think of “stars and bars” – a classic combinatorial trick.
- Always double‑check whether items are distinct or identical; it changes the formula.
Revision
Log in to practice.
0 views
0 suggestions