Solving the Odd 3-Digit Number Permutations

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SUMMARY

The discussion focuses on calculating the number of odd 3-digit numbers that can be formed using the digits 1 through 7, with each digit used only once. The key approach involves selecting three digits from the set and ensuring that the last digit is odd to satisfy the condition of the number being odd. The solution involves using combinatorial methods, specifically "Seven choose three," to determine the combinations of digits, followed by calculating the permutations of those combinations to find the total number of valid 3-digit numbers.

PREREQUISITES
  • Understanding of combinatorial mathematics, specifically combinations and permutations.
  • Familiarity with basic number properties, particularly odd and even numbers.
  • Knowledge of the concept of digit selection from a finite set.
  • Ability to apply factorial calculations for permutations.
NEXT STEPS
  • Study the concept of combinations, specifically "n choose k" calculations.
  • Learn about permutations and how to calculate them for a given set of items.
  • Explore the properties of odd and even numbers in number theory.
  • Practice solving similar combinatorial problems involving digit arrangements.
USEFUL FOR

Students studying combinatorial mathematics, educators teaching number theory, and anyone interested in solving mathematical puzzles involving digit permutations.

L²Cc
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Homework Statement


How many 3 digit numbers can be constructed from digits 1, 2, 3, 4, 5, 6, and 7 if each digit may be used once only and the number is odd?


2. The attempt at a solution
What number do they speak of? The resulting 3 digit number? How do I approach this equation?
 
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L²Cc said:

Homework Statement


How many 3 digit numbers can be constructed from digits 1, 2, 3, 4, 5, 6, and 7 if each digit may be used once only and the number is odd?


2. The attempt at a solution
What number do they speak of? The resulting 3 digit number? How do I approach this equation?

For example, I could pick out the numbers 1,2 and 3 and form the number

123

but also, I could form, 132 ,or 213, or 231, or 312, or 321.

But, that is just one way to pick three numbers (1,2,3). I could have chosen to pick out the numbers 3,5 and 1. And I could then form 6 different numbers (135,153,315,351,513,531) with those.

If I were you I would start off by thinking about how many different ways there are to choose three different things out of an array of 7 different things. "Seven choose three".

Then, you know that for any set of three, you can make 6 numbers, but you have to figure out how many of them are odd. Good luck.
 
L²Cc said:
What number do they speak of?


Oh. Yeah. They are probably talking about the *resulting* number (the three digit number). That is a very confusing way to word the problem. It is certainly vague.
 

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