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Permutation is a fundamental concept in Algebra and Combinatorics that deals with the arrangement of objects. When we want to know in how many different ways we can arrange a set of distinct elements, we use permutations.
A permutation of a set is an ordered arrangement of its elements. The order in which elements are arranged matters in permutations.
For example, consider the set \( {A, B, C} \). The permutations of 2 elements taken from this set are:
\( AB, AC, BA, BC, CA, CB \)
Notice that \( AB \neq BA \) because order matters.
If we want to arrange \( r \) objects out of \( n \) distinct objects, the number of permutations is given by:
\[P(n, r) = n \times (n-1) \times (n-2) \times \cdots \times (n-r+1)\]This can be compactly written using factorial notation as:
\[P(n, r) = \frac{n!}{(n-r)!}\]where \( n! = n \times (n-1) \times \cdots \times 2 \times 1 \) is the factorial of \( n \).
When all elements are distinct and each element can be used only once, permutations are without repetition.
Example: How many 3-letter words can be formed from the letters \( {A, B, C, D} \) without repeating any letter?
\[P(4, 3) = \frac{4!}{(4-3)!} = \frac{4 \times 3 \times 2 \times 1}{1!} = 24\]If repetition of elements is allowed, then for each position, we have \( n \) choices. For \( r \) positions, the total number of permutations is:
\[n^r\]Example: How many 3-digit numbers can be formed using digits \( {1, 2, 3} \) if repetition is allowed?
\[3^3 = 27\]When arranging \( n \) distinct objects in a circle, the number of distinct permutations is:
\[(n-1)!\]This is because in a circle, rotations are considered the same arrangement.
Example: How many ways can 5 friends sit around a round table?
\[(5-1)! = 4! = 24\]Many board exam questions test the understanding of permutations through arrangement and counting problems, including circular permutations and permutations with repetition. Understanding factorials and the permutation formula is essential to solve these problems quickly and accurately.
Problem: How many ways can 4 books be arranged on a shelf?
Solution:
Since all books are distinct and order matters, the number of arrangements is:
\[ 4! = 4 \times 3 \times 2 \times 1 = 24 \]Answer: 24 ways.
Problem: From 7 students, in how many ways can a president and a secretary be chosen?
Solution: Here, order matters because president and secretary are different roles.
\[ P(7, 2) = \frac{7!}{(7-2)!} = \frac{7 \times 6 \times 5!}{5!} = 7 \times 6 = 42 \]Answer: 42 ways.
Problem: How many 4-letter words can be formed from the letters \( {A, B, C} \) if letters can be repeated?
Solution: Each position can be filled with any of the 3 letters.
\[ 3^4 = 81 \]Answer: 81 words.
Problem: In how many ways can 6 people be seated around a round table?
Solution: Number of circular permutations is:
\[ (6-1)! = 5! = 120 \]Answer: 120 ways.
Problem: How many different 5-letter arrangements can be made from the letters of the word "LEVEL"?
Solution: The word "LEVEL" has 5 letters with L repeated twice and E repeated twice.
Number of permutations:
\[ \frac{5!}{2! \times 2!} = \frac{120}{2 \times 2} = \frac{120}{4} = 30 \]Answer: 30 different arrangements.
| Concept | Formula | Description |
|---|---|---|
| Factorial | \( n! = n \times (n-1) \times \cdots \times 1 \) | Product of all positive integers up to \( n \) |
| Permutation of \( r \) objects from \( n \) | \( P(n, r) = \frac{n!}{(n-r)!} \) | Number of ordered arrangements of \( r \) elements from \( n \) |
| Permutation with repetition | \( n^r \) | Number of arrangements when repetition allowed |
| Circular permutation | \( (n-1)! \) | Number of ways to arrange \( n \) objects in a circle |
| Permutation with identical objects | \( \frac{n!}{n_1! \times n_2! \times \cdots} \) | Number of permutations when some objects are repeated |
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