Permutations and Combinations

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your exam.

Permutations and Combinations — Key Facts for NABE (Pakistan)

• Permutation (Order matters): nPr = n!/(n-r)! — arrangements of r items from n
• Combination (Order doesn’t matter): nCr = n!/[r!(n-r)!] — selections of r items from n
• Fundamental Principle: Multiplication for “and” (sequential), Addition for “or” (alternative)
• n! = n × (n-1) × … × 3 × 2 × 1 | Special: 0! = 1
• ⚡ Exam tip: Permutation = Selection + Arrangement; Combination = Selection only

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🟡 Standard — Regular Study (2d–2mo)

Standard content for students with a few days to months.

Permutations and Combinations — NABE (Pakistan) Study Guide

Fundamental Principle of Counting

Multiplication Rule (AND): If a task can be done in m ways, and another task can be done in n ways, both tasks can be done in m × n ways.

Example: You have 3 shirts and 4 pants. How many outfits?

• 3 × 4 = 12 outfits

Addition Rule (OR): If a task can be done in m ways OR n ways (mutually exclusive), it can be done in m + n ways.

Example: Going to college via bus (2 routes) OR via rickshaw (3 routes)

• 2 + 3 = 5 routes

Factorial

Definition: n! = n × (n-1) × (n-2) × … × 3 × 2 × 1

Special Values:

• 0! = 1 (by definition)
• 1! = 1
• 2! = 2
• 3! = 6
• 4! = 24
• 5! = 120

Permutations

Definition: Arrangements where ORDER matters

Formula:

nPr = n! / (n - r)!

Example: How many ways to arrange 3 books from 5?

• 5P3 = 5!/(5-3)! = 5!/2! = 120/2 = 60 ways

Special Cases:

• nPn = n! (arranging all n items)
• nP0 = 1 (one way to arrange nothing)
• nP1 = n (choose 1 from n in n ways)

Combinations

Definition: Selections where ORDER doesn’t matter

Formula:

nCr = n! / [r!(n - r)!]

Example: Selecting 3 students from 10 for a team

• 10C3 = 10!/[3!(7!)] = (10 × 9 × 8)/(3 × 2 × 1) = 720/6 = 120 ways

Key Property: nCr = nC(n-r)

NABE Exam Pattern

Common question types:

1. Basic permutation/combination calculations
2. Word problems with distinct objects
3. Problems with identical objects
4. Circular arrangements
5. Arrangement around a circle

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🔴 Extended — Deep Study (3mo+)

Comprehensive coverage for students on a longer study timeline.

Permutations and Combinations — Comprehensive NABE (Pakistan) Notes

Detailed Theory

1. When to Use Permutation vs. Combination

Permutation — When order matters:

• Arranging people in seats
• Forming words from letters
• Selecting batting order
• Ranking contestants

Combination — When order doesn’t matter:

• Selecting committee members
• Choosing menu items
• Selecting team players
• Drawing cards from deck

Rule of Thumb: If you can rearrange the selected items and get the same result, it’s a combination. If rearrangement gives different result, it’s a permutation.

2. Factorial Properties

Product Form:

n! = n × (n-1)!

Cancellation Property:

n!/(n-1)! = n

Very Important:

(n+1)! = (n+1) × n!

3. Permutation Formula — Derivation

nPr = n!/(n-r)!

Proof:

• First position: n choices
• Second position: (n-1) choices
• Third position: (n-2) choices
• …
• r-th position: (n-r+1) choices
• By multiplication: n × (n-1) × (n-2) × … × (n-r+1) = n!/(n-r)!

4. Combination Formula — Derivation

nCr = n! / [r!(n-r)!]

Proof:

• Each combination of r items can be arranged in r! ways (permutations)
• nPr = nCr × r!
• Therefore: nCr = nPr/r! = n!/[(n-r)!r!]

Pascal’s Triangle Property:

nCr = n-1Cr-1 + n-1Cr

Example: 5C3 = 5C2 = 10

• 4C2 + 4C3 = 6 + 4 = 10 ✓

5. Circular Permutations

Arrangement around a circle: (n-1)!

Why (n-1)!?

• Fix one person to break circular symmetry
• Remaining (n-1) people can be arranged in (n-1)! ways

Clockwise vs Anti-clockwise:

• If clockwise and anti-clockwise arrangements are considered SAME: (n-1)!/2
• This applies when there’s no distinction between clockwise and counterclockwise

Example: 5 people around a circular table

• 4! = 24 arrangements (if direction matters)
• 4!/2 = 12 arrangements (if direction doesn’t matter)

6. Permutations with Repetition

When some objects are identical:

n!/(p! × q! × r!)  [where p, q, r are counts of identical objects]

Example: Arrange letters of “STATISTICS”

• S appears 3 times, T appears 3 times, I appears 2 times, A, C each 1
• Total letters = 10
• Arrangements = 10!/(3! × 3! × 2!) = 3628800/(6 × 6 × 2) = 50,400

For identical objects around a circle: Use Burnside’s lemma (advanced)

7. Combinations with Repetition (Stars and Bars)

Selecting r items from n types (repetition allowed):

(n + r - 1)C(r) = (n + r - 1)C(n - 1)

Example: In how many ways can you buy 5 fruits from mangoes, apples, and oranges?

• n = 3 types, r = 5 fruits
• Ways = (3 + 5 - 1)C5 = 7C5 = 7C2 = 21 ways

Formula Derivation (Stars and Bars):

• Represent r identical items as stars: ****
• Place (n-1) bars to divide into n groups
• Total positions = r + (n-1)
• Choose (n-1) positions for bars = (r+n-1)C(n-1)

8. Restricted Combinations and Permutations

n objects with restrictions:

Example 1: Number of 4-digit numbers with distinct digits and divisible by 5

• Last digit must be 0 or 5
• Case A: Ends in 0: First digit (1-9, not 0): 8 choices, remaining 2: 8×7 choices
• Case B: Ends in 5: First digit (1-9, not 5, not 0): 7 choices, remaining 2: 8×7 choices
• Total = 8×8×7 + 7×8×7 = 448 + 392 = 840

Example 2: Select 3 men and 2 women from 5 men and 4 women

• 5C3 × 4C2 = 10 × 6 = 60

9. Division into Groups

Unequal Groups:

• Select for Group 1: nCk
• Select for Group 2: n-k Cl
• And so on…

Equal Groups (advanced):

• Divide n items into r equal groups of n/r each
• Number of ways = n!/[(n/r)!]^r × r! [if groups are labeled]
• Number of ways = n!/[r! × (n/r)!]^r [if groups are unlabeled]

10. Key Formulas Summary

Situation	Formula
Permutation of n, r at a time	nPr = n!/(n-r)!
Combination of n, r at a time	nCr = n!/[r!(n-r)!]
Circular permutation	(n-1)!
All n objects (arranged)	n!
Identical objects	n!/[p!q!r!…]
nCr = nC(n-r)	Symmetry
nC0 = nCn = 1	Boundary
nCr + nCr-1 = n+1Cr	Pascal’s identity

11. Common Mistakes to Avoid

1. Order vs. No Order: Read problem carefully to determine
2. Factorial Overflow: nPr and nCr for large n — simplify first
3. Overcounting: Be careful with overlapping cases
4. Repetition: Check if objects are distinct or identical
5. Circular Arrangements: Remember (n-1)! not n!

Practice Questions for NABE

1. Evaluate: 6P3, 8C4, 10C6
2. How many 4-digit even numbers can be formed using digits 1,2,3,4,5 (no repetition)?
3. In how many ways can 5 boys and 4 girls be arranged so that no two girls are together?
4. From 6 men and 4 women, a committee of 5 is formed with at least 2 women. Find the number of ways.
5. In how many ways can the letters of “ENGINEERING” be arranged?

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