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 :

Since the five-digit number must be greater than 40000, the only options for the first digit are 5, 7, or 9. That leaves 3 remaining choices for the first digit. <br/><br/>Since the number has to be divisible by 5, the last digit must be 0 or 5. If the first digit is 5, the last digit can only be 0, since digits cannot be repeated. If the first digit is 7 or 9, the last digit can be 0 or 5, so there are 2 choices. <br/><br/>Five possible configurations for the first and last digits : <br/><br/>- 5xxx0 <br/><br/>- 7xxx0 <br/><br/>- 7xxx5 <br/><br/>- 9xxx0 <br/><br/>- 9xxx5 <br/><br/>For each of these configurations, there are three places in the middle that can be filled with the remaining 4 unused digits. <br/><br/>Since repetition isn't allowed, there are 4 options for the second digit, 3 for the third, and 2 for the fourth, which can be represented as ⁴P₃, or permutations of 4 items taken 3 at a time. This is equal to 4 $\times$ 3 $\times$ 2 = 24. <br/><br/>So, for each of the 5 configurations, there are 24 ways to arrange the middle three digits. Therefore, the total number of five-digit numbers that meet the criteria is 5 $\times$ 24 = 120.