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Permutation and Combination

Introduction

Permutation and Combination is a fundamental topic in combinatorics with wide applications in mathematics, computer science, probability, statistics, and real-life counting problems. A permutation refers to arrangements of objects where order matters, while a combination refers to selections of objects where order does not matter. Mastering these topics requires understanding factorials, counting principles, and standard formulas. We will cover everything from basic concepts to advanced theorems, solved examples, and applications.

Factorials and Their Properties

The factorial of a non-negative integer n is denoted by n! and defined as n! = n × (n−1) × (n−2) × ... × 3 × 2 × 1 with 0! = 1 by convention. Example: 5! = 5×4×3×2×1 = 120. Factorials grow very fast, e.g., 10! = 3,628,800.

Exponent of a Prime in n!: For a prime p, exponent in n! = [n/p] + [n/p²] + [n/p³] + ... where [ ] is the greatest integer function. Example: exponent of 3 in 100! = [100/3]+[100/9]+[100/27]+[100/81] = 33+11+3+1=48.

Trailing Zeros in n!: Trailing zeros come from factor 10 = 2×5. Since 2s are more frequent, count 5s: [n/5]+[n/25]+[n/125]+... Example: zeros in 100! = [100/5]+[100/25] = 20+4=24.

Basic Counting Principles

Addition Principle: If one task can be done in m ways and another in n ways and both cannot be done together, total ways = m+n. Example: Two-digit numbers ending in 7 or even digits → count separately then add.

Multiplication Principle: If a task A can be done in m ways and after that task B in n ways, total ways = m×n. Example: Toss a coin (2 ways) and roll a die (6 ways) → total = 2×6=12.

Permutations

Definition: The number of arrangements of n objects taken r at a time is nPr = n!/(n−r)!. Example: Arrange 3 books from 5 → 5P3 = 5!/(5−3)! = 5!/2! = 60.

When r=n → nPn = n!. When r=0 → nP0=1.

If repetition allowed → n^r arrangements. Example: 3-digit numbers using {1,2,3} with repetition → 3^3=27.

Restricted permutations:

• One object always included → fix it, arrange remaining.

• One object never included → exclude it.

• Two objects together → treat as single unit then multiply by arrangements within. Example: 5 boys and 5 girls in row with no two girls together → arrange boys first then place girls in gaps.

Circular permutations: n objects in circle → (n−1)!. If clockwise=anticlockwise same → (n−1)!/2.

Combinations

Definition: Number of selections of r objects from n without order = nCr = n!/(r!(n−r)!). Example: Choosing 3 students from 10 → 10C3=120.

Properties:

1. nCr = nC(n−r)

2. nCr + nC(r−1) = (n+1)Cr

3. Sum of all combinations from r=0 to n = 2^n

Examples:

1. Number of diagonals in n-sided polygon = nC2 − n = n(n−3)/2.

2. Handshakes among n people = nC2.

Mixed Problems

Example: Form 4-digit numbers from {1,2,3,4,5} without repetition → first choose 4 digits from 5: 5C4=5 then arrange: 4!=24 → total=5×24=120.

Example: Committee of 2 men and 3 women from 5 men 6 women → choose men:5C2=10, women:6C3=20, total=10×20=200.

Permutations and Combinations with Repetition

Combinations with repetition: Selecting r objects from n types with repetition = (n+r−1)Cr. Example: Number of non-negative integer solutions of x1+x2+...+xn=r = (n+r−1)Cr.

Multinomial Theorem

For positive integers n and x1,x2,...,xk: (x1+x2+...+xk)^n = n!/(n1!n2!...nk!) × x1^n1 x2^n2 ... xk^nk summed over n1+n2+...+nk=n.

Probability Applications

If all outcomes equally likely, P(E)=favorable outcomes/total outcomes. Example: Probability of getting 2 heads in 3 coin tosses → total=2^3=8, favorable=3C2=3 → P=3/8.

Example: Probability that 2 cards from 52 are both aces → favorable=4C2=6, total=52C2=1326 → P=6/1326=1/221.

Advanced Identities and Results

1. nCr × r = n × n−1C r−1

2. nPr = nCr × r!

3. nCr × mCk = number of ways choosing r from n and k from m.

4. Sum of r×nCr over r=0 to n = n×2^(n−1).

Derangements

Number of permutations with no object in its original place (D_n): D_n = n!(1−1/1!+1/2!−1/3!+...+ (−1)^n/ n!). Example: D3=3!(1−1+1/2)=3.

Partitions of Integers

Number of ways to write n as sum of positive integers disregarding order = partition function p(n). Example: p(4)=5 since 4=4,3+1,2+2,2+1+1,1+1+1+1.

Examples for Practice

1. How many 4-digit numbers with distinct digits? → First digit 1–9 →9, others any remaining digits →9×9×8×7=4536.

2. Probability of 2 boys and 2 girls in committee of 4 from 5 boys 5 girls → boys:5C2=10, girls:5C2=10, total:10×10=100, total committees:10C4=210 → P=100/210=10/21.

3. Number of words from letters of ‘MISSISSIPPI’ → total letters=11, frequencies: M=1,I=4,S=4,P=2 → 11!/(1!4!4!2!)=34650.

4. Circular arrangements of 6 persons where 2 always together → treat them as one → (5−1)!=4!=24 × arrangements within pair (2)!=2 → total=48.

More Advanced Examples

• Number of solutions of x1+x2+x3=10 in non-negative integers → (10+3−1)C(3−1)=12C2=66.

• Number of 5-digit numbers divisible by 5 from digits {0–9} without repetition → last digit 0 or 5, count separately, sum.

• Ways to seat n men n women alternately in row → n!×n!.

• Number of rectangles in m×n grid = mC2×nC2.