Permutations and Combinations

Permutations and combinations serve as cornerstone principles in Mathematics, particularly within the realm of combinatorics. Permutation and combination offer insight into the diverse methods of selecting and organizing elements from a given set. Permutations specifically concentrate on the arrangement of items, emphasizing the significance of order in selection. Conversely, combinations address the selection of items without regard to their order. These foundational concepts wield significant influence across various Mathematical domains.

The objective of this article is to offer a thorough grasp of permutations and combinations. It delves into their definitions, formulas, distinctions, and applications, and offers solved examples for clarity. Additionally, a Permutation and Combination Worksheet is provided to aid students in honing their comprehension and skills in these areas.

1.0What are Permutations and Combinations?

Definition of Permutation

Within the domain of mathematics, a permutation signifies the systematic arrangement of all elements within a set according to a defined order. When the elements of a set are reorganized, assuming they are initially ordered, this process is termed permuting. Permutations hold significance across diverse mathematical contexts, particularly in analyzing alternate orderings of finite sets.

Definition of Combination

A combination entails choosing items from a collection without considering the order of selection. In straightforward scenarios, determining the number of combinations is manageable. Technically, a combination involves selecting 'n' items taken 'k' at a time without repetition.

2.0Permutations and Combinations Formulas

Permutations and combinations are accompanied by several essential formulas. Among them, two fundamental formulas stand out:

Permutation Formula

In a permutation, 'r' items are chosen from a set of 'n' items, where the order of selection holds significance, and replacement is prohibited.

The formula for permutations is expressed as: 

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

Combination Formula

A combination involves the selection of 'r' items from a set of 'n' items, where the order of selection doesn't matter, and replacement is not allowed.

The formula for combinations is as follows:

${}^n{C}_r=\left(\genfrac{}{}{0ex}{}{n}{r}\right)=\frac{{}^n{P}_r}{r!}=\frac{n!}{r!(n-r)!}$

3.0Difference Between Permutations and Combinations

4.0Applications of Permutations and Combinations

• Permutations are commonly applied when handling datasets where the sequence holds significance. Conversely, combinations find utility in scenarios involving data groups where sequence is irrelevant.
• Combinatorial Analysis: Permutations and combinations are extensively used in combinatorial analysis to count the number of possible outcomes in various scenarios.
• Probability: They play a crucial role in probability theory, especially in calculating probabilities of events in experiments and games.
• Statistics: Permutations and combinations are utilized in statistical analysis for calculating permutations of data samples and combinations of variables in experiments.
• Games and Puzzles: Permutations and combinations are commonly used in designing puzzles, games, and recreational mathematics problems, where players need to arrange or select elements in specific ways to achieve a desired outcome.
• Design and Analysis: In design theory and analysis, permutations and combinations are applied to create optimal designs for experiments and to analyze the effectiveness of various treatments or factors.

5.0Permutations and Combinations Solved Examples

Example 1: In how many ways can 5 persons be made to occupy three different chairs.

Solution:  

${}^5{P}_3=5\times 4\times 3$ =60

Example 2: Find the value of { }^5 C_2 

Solution:

${}^5{C}_2=\frac{\underline{5}}{\underline{5-2}\cdot [2}=\frac{5.4}{2}$ =10 

Example 3: Find the value of n such that ⁿP₅ = 42 ⁿP₃, n > 4 

Solution:

Given that ⁿP₅ = 42 ⁿP₃

or n (n – 1) (n – 2) (n – 3) (n – 4) = 42 n (n – 1) (n – 2)

Given n > 4 so n (n – 1) (n - 2) ≠ 0

Therefore, dividing both sides by n (n – 1) (n – 2), we get-

(n – 3) (n – 4) = 42 

or n² – 7n–30=0 

or n² – 10n + 3n – 30 = 0

or (n – 10) (n + 3) = 0

or n – 10 = 0 or n + 3 = 0

or n = 10 or n = –3

Since n cannot be negative, so n = 10.

Example 4: Find r, if 5 ⁴P_r = 6 ⁵P_r–1.

Solution:

We have 5 ⁴P_r = 6 ⁵P_r-1

Or $5\times \frac{4!}{(4-r)!}=6\times \frac{5!}{(5-r+1)!}$

Or $\frac{5!}{(4-r)!}=\frac{6\times 5!}{(5-r+1)(5-r)(5-r-1)!}$

or (6 – r) (5 – r) = 6

or r² – 11r + 24 = 0

or r² – 8r – 3r + 24 = 0

or (r – 8) (r – 3) = 0

or r = 8 or r = 3.

Hence r = 8.

Example 5: ${}^r{C}_r+{}^{r+1}{C}_r+{}^{r+2}{C}_r+\dots \dots \dots +{}^n{C}_r=\text{ ? }$

Solution:

${}^r{C}_r+{}^{r+1}{C}_r+{}^{r+2}{C}_r+\dots \dots \dots +{}^n{C}_r$

$={}^{r+1}{C}_{r+1}+{}^{r+1}{C}_r+{}^{r+2}{C}_r+\dots \dots \dots .+{}^n{C}_r\left[{}^r{C}_r={}^{r+1}{C}_{r+1}=1\right]$

$={}^{r+2}{C}_{r+1}+{}^{r+2}{C}_r+\dots \dots \dots .+{}^n{C}_r$

$={}^{r+3}{C}_{r+1}+{}^{r+3}{C}_r+\dots \dots \dots +{}^n{C}_r$

$={}^n{C}_{r+1}+{}^n{C}_r={}^{n+1}{C}_{r+1}={}^{n+1}{C}_{n-r}$

Example 6: A committee of 5 people is to be formed from a group of 8 women and 6 men. In how many ways can this be done if the committee must contain at least 3 women?

Solution:

1. Selecting 3 women and 2 men: This can be done in ⁸C₃ × ⁶C₂ ways.

2. Selecting 4 women and 1 man: This can be done in ⁸C₄ × ⁶C₁ ways.

3. Selecting 5 women and 0 men: This can be done in ⁸C₅ × ⁶C₀ ways.

The total number of ways to form the committee is the sum of these three cases.

= ⁸C₃ × ⁶C₂ + ⁸C₄ × ⁶C₁ + ⁸C₅ × ⁶C₀ 

= 840 + 420 + 56

= 1316

6.0Permutations and Combinations Practice Questions

1. A man has to travel from Bombay to Goa via Pune. He has 3 ways from Bombay to Pune & 4 ways from Pune to Goa. In how many ways he can plan a return journey from Bombay to Goa back to Bombay?

(A) 121 (B) 144 (C) 169 (D) 289

2.  The number of 3 digits odd numbers formed using the given digits 1, 2, 3, 4, 5, 6 when repetition is not allowed, is

(A) 60 (B) 108 (C) 36 (D) 30

3. How many five digit numbers can be made using 0, 1, 2, 4, 5, 3 which are divisible by 3. (If repetition not allowed)

(A) 186 (B) 196 (C) 206 (D) 216

7.0Solved Questions on Permutations and Combinations

1. What is permutation?

Ans: A permutation refers to the arrangement of elements in a specific order.

2. When is permutation used?

Ans: Permutations are used when the order of selection matters, such as arranging items or selecting leaders in a sequence.

3. What is a combination?

Ans: A combination involves selecting items from a collection without considering the order.

4. When is a combination used?

Ans: Combinations are used when the order of selection is irrelevant, like choosing a committee or selecting items for a group.

5. What are the formulas for permutation and combination?

Ans: The permutation formula is ${}^{\mathrm{n}}{P}_r=\frac{n!}{(n-r)!}$; The combination formula is ${}^n{C}_r=\left(\genfrac{}{}{0ex}{}{n}{r}\right)=\frac{{}^n{P}_r}{r!}=\frac{n!}{r!(n-r)!}.$

6. Can you explain the difference between permutations and combinations?

Ans: Permutations involve arranging items in a specific order, while combinations focus on selecting items without regard to order.

7. How do I calculate permutations and combinations?

Ans: Use the permutation formula for ordered arrangements and the combination formula for unordered selections.

8. Are there any practical applications of permutations and combinations?

Ans: Yes, permutation and combination concepts are used in fields like probability, statistics, cryptography, and computer science.

9. Can permutations and combinations be used interchangeably?

Ans: No, they represent distinct concepts with different mathematical implications.