1. ## permutation!

How many 3-digit odd numbers can be formed with the digits 1, 2, 3, . . . , 9, if no digit is repeated in any
number?

pls help on the problem above, thanks.

2. ## Re: permutation!

To get a number ending in 1- first can be any of 8 and second any of 7. So there are 8x7=56. But it will be the same for numbers ending in 3,5,7 or 9.

3. ## Re: permutation!

Can you think of a sequence of steps to construct such a number? E.g., "at first I choose the last digit, then ..." Try to calculate in how many ways each step can be done. Then usually you have to multiply the numbers for each step. It is important that all possible numbers can be constructed using these steps, and no number is constructed twice using different choices.

4. ## Re: permutation!

Surely that is exactly what I have done.

5. ## Re: permutation!

Originally Posted by biffboy
Surely that is exactly what I have done.
Yes. The main reason I made a post is that I did not see your post. Also, I wanted to give a hint rather than a solution.

6. ## Re: permutation!

Originally Posted by lawochekel
How many 3-digit odd numbers can be formed with the digits 1, 2, 3, . . . , 9, if no digit is repeated in any
number?

pls help on the problem above, thanks.
$N=5 \cdot \frac{8!}{(8-2)!}=5 \cdot 8 \cdot 7 =280$