Solving the Odd 3-Digit Number Permutations

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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.