Answer:
250
Step-by-step explanation:
Given
[tex]n =3[/tex] --- digit code
[tex]First = Even[/tex]
[tex]Last=Odd[/tex]
Required
The different combination of codes
The set of even number is:
[tex]Even = \{0,2,4,6,8\}[/tex]
The set of odd is:
[tex]Odd = \{1,3,5,7,9\}[/tex]
This means that:
The first can be any of the 5 even digits
The last can be any of the 5 odd digits
Since there is no restriction in selection;
The [tex]second[/tex] can be [tex]any[/tex] of [tex]the[/tex] 10 digits
So, the possible combination (k) is:
[tex]k =5 * 10 * 5[/tex]
[tex]k =250[/tex]