Permutation and Combination – Quick Revision Notes

---

1. Basic Terms

🔹 Permutation

Arrangement of objects in a specific order.
👉 Order matters.

Example:
Arranging A, B, C → ABC, ACB, BAC, BCA, CAB, CBA → 6 ways.

Formula:

$${}^n{P}_r=\frac{n!}{(n-r)!}$$

where

• n = total number of items

• r = number of items selected

• n! = factorial of n = n × (n−1) × (n−2) × … × 1

---

🔹 Combination

Selection of objects without caring about order.
👉 Order doesn’t matter.

Example:
Choosing 2 from A, B, C → AB, AC, BC → 3 ways.

Formula:

$${}^n{C}_r=\frac{n!}{r!(n-r)!}$$

---

2. Relation Between P and C

$${}^n{P}_r={}^n{C}_r\times r!$$

---

3. Basic Results

Type	Formula
Total permutations of n objects	n!
Circular permutation	(n − 1)!
Circular permutation (if clockwise = anticlockwise same)	(n − 1)! / 2
Number of ways to arrange n distinct items, with p alike of one kind, q alike of another, etc.	n! / (p! × q! × …)
Number of selections	nCr

---

4. Important Variations

🔸 When Repetition Is Allowed

• Permutation (repetition allowed):

  $$n^r$$
  Example: Number of 3-digit numbers formed from digits 1–5 (repetition allowed) = 5³ = 125

• Combination (repetition allowed):

  $${}^{n+r-1}{C}_{\mathrm{r}}$$ Example: Selecting 3 fruits from 5 types (repetition allowed) → C(7,3) = 35 ways

---

🔸 When All Items Are Used

Total arrangements of n distinct objects = n!

Example: 5 letters → 5! = 120 arrangements.

---

5. Circular Permutations

Case	Formula
n distinct objects around a circle	(n − 1)!
n distinct objects around a circle (rotation & reflection same)	(n − 1)! / 2
Arranging n people in a row (linear)	n!
Arranging n people around a round table	(n − 1)!

Example: 4 people around a round table = (4−1)! = 6 ways.

---

6. Conditional Permutations

Condition	Formula / Explanation
If certain objects must always be together	Treat them as one unit. Then multiply by internal arrangements.
If certain objects must never be together	Total – (No. of ways they are together).
Arranging vowels/consonants separately	Use grouping method.

---

Example:

How many words from “DELHI” so that vowels are always together?
Vowels: E, I → treat as one (EI) → units: (EI), D, L, H → 4 units → 4! = 24 ways
Internal arrangement of (EI) → 2! = 2 ways
Total = 24 × 2 = 48 ways

---

7. Problems on Digits or Numbers

Type	Approach
Forming 3-digit numbers using given digits	Use nPr (for arrangement).
Repetition allowed	n^r
Even/Odd restriction	Restrict last digit first.
Distinct digits only	Use nPr after restriction.

---

Example:

Form 3-digit even numbers using digits 1–5, no repetition.
Last digit (even): 2 or 4 → 2 ways
Remaining 2 digits = 4P2 = 12 ways
Total = 2 × 12 = 24 numbers

---

8. Combination in Real-Life Problems

Problem Type	Formula / Concept
Selecting a team of r from n persons	nCr
Selecting committee with conditions	Subtract restricted cases
Choosing 2 or more from group	Sum of combinations (nC2 + nC3 + …)
Handshake problem	nC2

Handshake Example:
If 10 persons shake hands once → 10C2 = 45 handshakes.

---

9. Mixed Formula Summary

Concept	Formula
nPr	n! / (n − r)!
nCr	n! / [r!(n − r)!]
nPr = nCr × r!	Relation
Circular permutation	(n − 1)!
If identical objects exist	n! / (p! × q! × …)
Repetition allowed	n^r
Combination with repetition	(n + r − 1)Cr

---

10. Key Tricks for Exams

✅ If order matters → Permutation
✅ If order doesn’t matter → Combination
✅ Circular arrangement → (n−1)!
✅ Repetition allowed → Use power formula
✅ “At least” type → Use total − restricted
✅ For vowels/consonants → Grouping helps

---

11. Practice Examples

1️⃣ How many 4-letter words can be formed from “APPLE”?
→ 5 letters with 2 P’s → 5! / 2! = 60

2️⃣ How many ways can 5 people sit around a round table?
→ (5 − 1)! = 24

3️⃣ In how many ways can a committee of 3 be selected from 6?
→ 6C3 = 20

4️⃣ How many 3-digit numbers from digits 1–6 (no repetition)?
→ 6P3 = 120

---

✅ Quick Recap Table

Case	Formula	Example
Arrangement (Order matters)	nPr	6P3 = 120
Selection (Order doesn’t matter)	nCr	6C3 = 20
Repetition allowed	n^r	5³ = 125
Circular arrangement	(n−1)!	5 → 24
Identical objects	n!/(p!q!…)	APPLE → 60