To solve this question, we need to consider that the password must end with an even number. From the available digits, we have 3 even numbers (4, 8) and 4 odd numbers (1, 3, 5, 7, 9). The last digit of the password must be even, so we have 2 options. For the other three digits, we can choose any of the remaining 6 digits, without repetition. So, we have 6 options for the first digit, 5 for the second, and 4 for the third. Multiplying all these options, we get 2*6*5*4 = 240 possible passwords.