Permutation and Combination Problems Made Simple for Students

Permutation and combination problems concern the enumeration of arrangements and selections from finite sets, utilizing precise mathematical definitions and combinatorial formulas.

Formal Definitions and Notation in Permutation and Combination Problems

Definition: A permutation of $n$ distinct objects taken $r$ at a time is an ordered arrangement, whereas a combination is an unordered selection of $r$ objects from $n$.

The number of permutations of $n$ objects taken $r$ at a time is denoted as $_nP_r = \dfrac{n!}{(n - r)!}$, valid for $0 \leq r \leq n$.

The number of combinations is denoted as $_nC_r = \dfrac{n!}{r!(n - r)!}$, where $n!$ represents the factorial of $n$.

The value $0!$ is defined as $1$ by convention to ensure continuity of the factorial function.

Relationships and Basic Identities Used in Enumeration

Basic relationships underpinning these problems include $_nP_r = {_nC_r} \cdot r!$, and the symmetric property $_nC_r = {_nC_{n - r}}$ for any $0 \leq r \leq n$.

The Pascal's Rule for combinations states that $_nC_r + {_nC_{r - 1}} = {_n+1C_r}$, providing a recursive means to compute values.

For selection among indistinguishable objects, the multinomial coefficient $\dfrac{n!}{n_1!n_2!\cdots n_k!}$ is used, where $n_1+n_2+\cdots+n_k=n$ partitions the objects into groups of similar elements.

For arrangements with repetition permitted, the number of $r$-length strings from an $n$-element set is $n^r$.

Canonical Exam Patterns and Problem Structuring in JEE

Permutations are commonly assessed via arrangements of digits, letters, or persons, often applying constraints such as adjacency or fixed positions.

Combinations are typically examined in scenarios involving committee formation, selection of groups, or count of possible pairings, sometimes with restrictions on group composition.

Mixed situations requiring sequential use of both principles—such as selection followed by arrangement—are common in higher-difficulty questions.

Evaluation of Expressions Involving Factorials and Binomial Coefficients

Example: Evaluate $6!$.

Substitution gives $6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1$, which simplifies to $720$.

Example: Compute $_7C_3$.

Substitute into the formula to get $_7C_3 = \dfrac{7!}{3! \times 4!}$.

Writing $7! = 5040$, $3! = 6$, and $4! = 24$, the expression becomes $\dfrac{5040}{6 \times 24}$.

The denominator is $144$, so the result is $35$.

Problem Solving Using Permutation Principles

Example: In how many ways can the letters of the word "FIGHT" be arranged?

Since there are $5$ distinct letters, the number of arrangements is $5!$.

Write $5! = 120$.

Example: Find the number of distinct arrangements of "GEOGRAPHY" in which the vowels are always together.

The vowels $E$, $O$, $A$ are treated as a unit. Therefore, consider $7$ entities: $G$, $G$, $R$, $P$, $H$, $Y$, and the unit of vowels.

Arrangements of these $7$ units, accounting for repeated $G$'s, is $\dfrac{7!}{2!} = 2520$.

The $3$ vowels can be arranged within their unit in $3! = 6$ ways.

Thus, the total number of arrangements is $2520 \times 6 = 15120$.

Allocation and Pair Formation via Combination Counting

Example: If $12$ persons each shake hands with every other person, how many handshakes occur?

Each handshake pairs $2$ individuals; the number of unordered pairs is $_{12}C_2$.

Substitute: $_{12}C_2 = \dfrac{12 \times 11}{2} = 66$.

Example: A committee of $5$ is formed from $7$ men and $6$ women, with at least $3$ men required. In how many ways can this be done?

Consider cases: $(3\text{ men},2\text{ women})$, $(4,1)$, $(5,0)$.

Total ways: $7C3 \times 6C2 + 7C4 \times 6C1 + 7C5$.

Compute: $35\times 15 + 35\times 6 + 21 = 525+210+21 = 756$.

Common Constraints and Typical Errors in Permutation-Combination Problems

Common restrictions involve placements (e.g., fixed odd/even positions), imposed adjacency or separation, and repetition decisions.

Common Error: Confusing arrangement (permutation) with selection (combination), especially in problems that require both stages, leads to misapplication of formulas.

Problems with repeated elements require division by the factorials of repeated counts, as in the arrangement of words with identical letters; omission results in overcounting.

For additional foundational material, refer to Permutation And Combination Overview.

Permutation and Combination in Probability and Set Problems

Enumeration of cases forms the basis for calculating probabilities or counting subsets, with $_nC_r$ providing the number of $r$-element subsets of an $n$-element set.

To deepen understanding of advanced combinatorial structures and alike object arrangements, see Permutation And Combination Of Alike Objects.

• Permutation formula for distinct objects
• Combination formula and symmetric properties
• Arrangements with identical elements
• Counting with and without repetition
• Committee and group formation strategies
• Constraints on arrangement (adjacency, position)
• Errors arising from order vs. selection distinction
• Recursive computation techniques

For JEE-level problems mixing both arrangement and selection constraints, consult Problems On Permutation And Combination.

Systematic mastery requires frequent practice. Refer to Important Questions On Permutations And Combinations for further exam-oriented problems.

To access dedicated revision tools and theory notes, visit Permutations And Combinations Revision Notes.