The number of five digit numbers, greater than 40000 and divisible by 5, which can be formed using the digits 0,1,3,5,7 and 9 without repetition, is equal to 

$\mathrm{Number}\mathrm{of}\mathrm{ways}$
$\mathrm{Case}1:5\times \times \times 0 4\times 3\times 2=24$
$\mathrm{Case}2:7\times \times \times 0 4\times 3\times 2=24$
$\mathrm{Case}3:\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}7\hspace{0.17em}\hspace{0.17em}\times \times \times \hspace{0.17em}\hspace{0.17em}5 4\times 3\times 2=24$
$\mathrm{Case}4:9\times \times \times 0 4\times 3\times 2=24$
$\mathrm{Case}5:9\times \times \times 5 4\times 3\times 2=24$
$\therefore \mathrm{Total}\mathrm{Number}\mathrm{of}\mathrm{ways}=120$