Number of distinct permutations: - Appcentric
Number of Distinct Permutations: A Complete Guide
Number of Distinct Permutations: A Complete Guide
When working with permutations, one fundamental question arises: how many distinct ways can a set of items be arranged? Understanding the number of distinct permutations is essential in mathematics, computer science, statistics, and real-world applications like cryptography and combinatorics. This article explores the concept of distinct permutations, how to calculate them, and real-world implications.
What Are Distinct Permutations?
Understanding the Context
A permutation refers to an arrangement of all or part of a set of items where the order matters. A distinct permutation considers unique sequences when repeating elements are present. For example, the string “AAB” has fewer distinct permutations than “ABC” due to the repetition of the letter ‘A’.
How to Calculate the Number of Distinct Permutations
1. Permutations of Distinct Objects
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Key Insights
If you have n distinct items, the total number of permutations is simply:
\[
n! = n \ imes (n-1) \ imes (n-2) \ imes \dots \ imes 1
\]
For example, “ABC” has \( 3! = 6 \) permutations: ABC, ACB, BAC, BCA, CAB, CBA.
2. Permutations with Repeated Items
When items are repeated, the formula adjusts by dividing by the factorial of the counts of each repeated item to eliminate indistinguishable arrangements.
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Use modular arithmetic: $ 2^3 = 8 \equiv 1 \mod 7 $, so $ 2^3 \equiv 1 \Rightarrow 2^{10} = (2^3)^3 \cdot 2 = 1^3 \cdot 2 = 2 \mod 7 $. Thus, the remainder is $ \boxed{2} $.Question: What is the smallest four-digit number that is divisible by the least common multiple of 12 and 18, and also leaves a remainder of 3 when divided by 7? To find the smallest four-digit number satisfying the given conditions, we proceed in steps.Final Thoughts
If a word or set contains:
- \( n \) total items
- \( n_1 \) identical items of type 1
- \( n_2 \) identical items of type 2
- …
- \( n_k \) identical items of type k
where \( n_1 + n_2 + \dots + n_k = n \), then the number of distinct permutations is:
\[
\frac{n!}{n_1! \ imes n_2! \ imes \dots \ imes n_k!}
\]
Example:
How many distinct permutations of the word “BANANA”?
Letters: B, A, N, A, N, A
Counts:
- 1 A
- 3 Ns
- 1 B
Total letters: \( n = 6 \)
\[
\ ext{Distinct permutations} = \frac{6!}{3! \ imes 1! \ imes 1!} = \frac{720}{6 \ imes 1 \ imes 1} = 120
\]
