Hey everyone. Please help me if you can to solve this problem. It is either combination or permutation, I'm not sure, but here is the problem itself:

How many 3-digit positive integers are odd and do not contain the digit "5".

Amanda

2. Hello, Snowing Amanda!

How many 3-digit positive integers are odd and do not contain the digit "5"?
Since the order of the digits is important, these are permutations.

We have a three-digit number: .$\displaystyle \boxed{X}\;\boxed{Y}\;\boxed{Z}$

The first digit, $\displaystyle X$, can be $\displaystyle \{1,2,3,4,6,7,8,9\}$ . . . 8 choices.

The middle digit, $\displaystyle Y$, can be $\displaystyle \{0,1,2,3,4,6,7,8,9\}$ . . . 9 choices.

The last digit, $\displaystyle Z$, can be $\displaystyle \{1,3,7,9\}$ . . . 4 choices.

Therefore, there are: .$\displaystyle 8\cdot9\cdot4 \:=\:288$ such numbers.

3. ## Tnanks so much!!!

Oh, thank you so very much! I've tried to solve it with all possible formulas, and never got it right. and with your answer it made it so easy for me!
thanks again!!!