When should I use the Beggar's Method versus basic combinations ${}^n C_r$?

Use ${}^n C_r$ when selecting $r$ distinct items out of $n$ distinct options where each item can be picked at most once. Use the Beggar's Method (${}^{n+r-1} C_{r-1}$) when distributing $n$ *identical* objects into $r$ *distinct* categories or bins, allowing individual bins to hold multiple items.

How do I avoid overcounting in group division problems?

When dividing $n$ distinct objects into groups of equal sizes (for instance, dividing $6$ items into $3$ groups of $2$ items each), dividing by $m!$ (where $m$ is the number of equal-sized groups) is mandatory to eliminate uncounted permutations between identical group sizes. The general formula is $\frac{n!}{(k!)^m m!}$.

What is the key strategy for solving circular permutation problems with restrictions?

Always seat the unconstrained or reference object first to break the circular symmetry. This converts the circular problem into a standard linear arrangement of size $(n-1)$. Afterwards, apply the standard linear gap method or string method for the remaining constrained objects.

Permutations and Combinations Notes, Formulas, Traps & PYQs | ExamiFly

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Solved High-Yield Sample Questions

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Question #1JEE Main 2024

The total number of $3$ -letter words that can be formed using the letters of the word $\text{'EXAMINATION'}$ is:

Solution Locked

Step-by-Step Solution & Concept

Step 1

Analyze letter frequencies in $\text{'EXAMINATION'}$ . There are $11$ total letters comprising $8$ distinct types: $3$ pairs $\{A, I, N\}$ and $5$ singletons $\{E, X, M, T, O\}$ .

Step 2

Calculate Case 1 (All $3$ letters distinct): Choose $3$ types out of $8$ available types and arrange them: ${}^8 C_3 \times 3! = 56 \times 6 = 336$ .

Step 3

Calculate Case 2 ( $2$ identical letters, $1$ distinct): Choose $1$ pair from $3$ available pairs, $1$ distinct letter from the remaining $7$ types, and arrange them: ${}^3 C_1 \times {}^7 C_1 \times \frac{3!}{2!} = 3 \times 7 \times 3 = 63$ . Summing both cases yields $336 + 63 = 399$ .

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Question #2JEE Main 2023

The number of non-negative integer solutions to the equation $x_1 + x_2 + x_3 + x_4 = 20$ such that $x_1 \ge 1$ , $x_2 \ge 2$ , $x_3 \ge 3$ , and $x_4 \ge 4$ is:

Solution Locked

Step-by-Step Solution & Concept

Step 1

Transform the constrained variables $x_1, x_2, x_3, x_4$ into new non-negative variables $y_i \ge 0$ using substitutions: $y_1 = x_1 - 1$ , $y_2 = x_2 - 2$ , $y_3 = x_3 - 3$ , and $y_4 = x_4 - 4$ .

Step 2

Substitute these into the main equation: $(y_1 + 1) + (y_2 + 2) + (y_3 + 3) + (y_4 + 4) = 20$ , which simplifies directly to $y_1 + y_2 + y_3 + y_4 = 10$ .

Step 3

Apply the Beggar's Method formula ${}^{n+r-1}C_{r-1}$ with $n = 10$ identical units and $r = 4$ variables, obtaining ${}^{10+4-1}C_{4-1} = {}^{13}C_3 = \frac{13 \times 12 \times 11}{6} = 286$ .

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Mastering Permutations and Combinations: Formulas, Traps & Exam Strategy | ExamiFly

Chapter Overview

Core Subtopics & Syllabus Breakdown

Key Formulas & High-Yield Equations

Permutations of Distinct Objects

Combinations of Distinct Objects

Permutations of Objects with Repetition (Multiset)

Circular Permutations

Distribution of Identical Objects into Distinct Bins (Beggar's Method)

Derangements Formula

Common Mistakes & Traps to Avoid

Past Year Question (PYQ) Trends

Solved High-Yield Sample Questions

Frequently Asked Questions

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