How many 3 digit odd numbers greater than 600 can be formed using the digits(2,3,4,5,6 and 7)? In my worksheet the answer is 20 but i keep getting a different answer
 A: The first digit is either 6 or 7.
If it is 6 the the last digit is either 3, 5, 7.  The middle digit could be any of the remaining 4 not used.  That's 3*4 = 12 possibilities.
If the first digit is 7, the last digit is either 3, or 5.  The middle digit could be any of the remaining 4 not used.  That's 2*4 =8.
So there are 12 + 8 = 20 possibilities.
A: Since the answer is $20$, you aren't allowed to repeat digits.
If you aren't allowed to repeat digits then first count how many numbers start with $6$ and how many start with $7$. There are numerous ways to do this. One way is to separate whether the middle digit is even or odd. 
Number of numbers starting with 6 and having even middle digit is $3\times 3=9$ (i.e. $3$ choices of middle digit and $3$ choices of last digit since it must be odd). The number of numbers starting with 6 and having odd middle digit is $3\times 2=6$ (i.e. $3$ choices of middle digit and then only $2$ choices for the final digit since you used up one of the odds on the middle).
Similarly you get $3$ ways to make numbers starting with $7$ and having even middle digit and $2$ ways if they have odd middle digit.
This gives a total of $9+6+3+2=20$ numbers. If you are allowed to repeat digits then you just have $2\times 6 \times 3=36$ numbers (i.e. $2$ choices for first digit, $6$ choices for second, and $3$ choices for third).
A: Hint #1: You can only use $3$, $5$, or $7$ for the unit's digit, since the number has to be odd.
Hint #2: You can only use $6$ or $7$ for the hundred's digit, since the number has to be greater than $600$.
Hint #3: Can you use the same digit more than once? e.g. Does $667$ qualify for your requirements?
Now use the Product Rule of Counting, and the Addition Rule of Counting, if necessary.
Can you take it from here?
