'''We apply the first one when expressing <math>f(n-1)</math> - that is, if <math>n-2</math> positions are know, then if the <math>n-2</math> position is odd, indeed you have <math>10</math> ways to fill the <math>n-1^{th}</math> position, and if it's even you have 5 ways.
'''We apply the second one when expressing <math>f^{(odd)}(n-1)</math> - that is, if the <math>n-1</math> position is known and odd (which you have 5 options for) then you have <math>f(n-2)</math> ways to fill in the <math>n-2</math> previous positionsfor each of the 5 odd digits: <math>5f(n-2)</math>.
'''Adi