Question from Ryu:
Using only the digits 0 1 2 3 4 5 6 7 8 9 without repetition, how many integers are:
(a) Greater than 999, less than 10,000, start with an even digit and are even?
(b) Greater than 999, less than 10,000, start with an odd digit and are odd?
(c) Greater than 999, less than 10,000, start with an even digit and are odd?
(d) Greater than 999, less than 10,000, start with an odd digit and are even?


Hi,
I'm going to solve a similar problem and you use my solution to help you solve your problems.
Using only the digits 0 1 2 3 4 5 6 7 8 9 without repetition, how many integers are:
Greater than 99, less than 1,000, start with a digit which is divisible by 3, and are divisible by 5?
The integers you want don't have any repeating digits. Let's look at the other three conditions and see what they mean.
 Greater than 99, and less than 1,000
 The smallest integer greater than 99 is 100 and the largest integer less than 1,000 is 999. Thus the integers you want are between 100 and 999 so they are three digit numbers.
 Start with a digit which is divisible by 3.
 The digits divisible by 3 are 0, 3, 6, and 9.
 Divisible by 5.
 The integers which are divisible by 5 are precisely the integer with units digit either 0 or 5.
Since the number you want are 3 digit numbers the hundreds digit can't be 0. Thus the hundreds digit is 3, 6 or 9. Hence you have 3 choices for the hundreds digit.
The units digit must be 0 or 5 so there are 2 choices for the units digit. Hence you have $3 \times 2 = 6$ possible ways to choose the hundreds and units digit.
Whichever of the six pairs you choose you have 8 choices for the tens digit and hence you have $6 \times 8 = 48$ integers that satisfy the conditions in the problem.
I hope this helps,
Penny
