I am having a hard time with this question Any explanations

I am having a hard time with this question. Any explanations with the answers are greatly appreciated.

Solve the following problems. Hand write your solutions and explanations on your own paper. Do not write answers on this page. Show your work AND explain your reasoning using complete English sentences. Using the digits 0 through 9, find out how many 4-digit numbers can be configured based on the stated conditions: The number cannot start with zero and no digits can be repeated. The number must begin and end with an odd digit. (Repeated digits are okay.) The number must be at least 5000 and be divisible by 10. (Repeated digits are okay.) The number must be less than 3000 and must be even. No digits may be repeated in the last 3 digits. (That is, 2234 would be okay, but 2334 would not be okay.)

Solution

let the 4 digit number be _ _ _ _ or a b c d

1) a cannot be 0 and a,b,c,d- all different
ANS
So, in this case a can be any of numbers 1-9
so, ways for a=9
similarly as no repetition,
ways for b=9 ( as in b,c,d we can use digit 0 once)
ways for c=9-1=8
ways for d=8-1=7
total ways=9*9*8*7 = 4536

2) a and d are odd numbers= 1,3,5,7,9
ANS
So, in this case a and d can be any of numbers 1,3,5,7,9
so, ways for a=5
similarly as repetition,
ways for b=10
ways for c=10
ways for d=5
total ways=5*10*10*5= 2500

3) number>=5000 and divisible by 10
So, in this case a can be any of numbers 5-9 and for number to be divisible by 10 it must have atleast one 0 in units place
so, ways for a=5
similarly as repetition,
ways for b=10
ways for c=10
ways for d=1
total ways=5*10*10*1 = 500

4) number <=3000 and even -> d=2,4,6,8,0 and b,c,d cannot have any repetition.
So, in this case a can be any of numbers 1,2 and d=2,4,6,8,0 and b,c,d-no repetition
so, ways for a=2
similarly as no repetition in b,c,d->
ways for b=9
ways for c=8
ways for d=5
total ways=2*9*8*5 = 720