Solve this mathematical problem please

It's not a difficult one, so please give it a shot. Here it goes (the translation is kinda rough) :

How many numbers are there that follow these rules:
- the number has to have 5 figures,
- all the figures in the number have to be different one from another,
- all the figures in the number have to be either even or odd.

Thank you in regards.

Easy question when you reason it out. It helps to know a bit about probabilities.

Five digits, all different, all either even or odd.

Let 's start with the. "all odd" case. The first digit can be 1, 3, 5, 7, or 9, giving us five possibilities.

It doesn't matter which one you start with, there are 4 choices left for the next digit. Pick the second digit, there are now 3 choices left for the third digit. After the third digit is chosen, there are 2 choices left for the fourth digit. When the 4th digit is chosen, there is only once digit left. So, the number of all-odd digit numerals with five different numerals, is 5 x 4 x 3 x 2 x1 = 120.

For the even numeral case, we have to know if 0 is allowed as the first digit. If so, the even numeral case is the same as the odd numeral case. If not, go back to step one, but this time, there are only 4 possible first doigits, but there are stil 4 possible 2nd digits. This time, instead of 5x4x3x2x1, you have 4x4x3x2x1 = 96.

Add the number of odd numbers to the number of even numbers, and you have your answer.

for diff. you have to choose one of 0,1,2,3,4,5,6,7,8,9 10 but you cannot choose 0 1st so we have to from 9 digits 2nt 9 digits 3rd 8 4th 7 and 5th 6 9x9x8x7x6=----

even 0,2,,4,6,8 1st 4 digits 2nt 4 digits 3rd 3 digits 4th 2 digits 5th1 4x4x3x2x1=96

odd 1,3,5,7,9 5! =240