Permutations of numbers 
Given the five digits $1,3,4,6,$ and $7$. In the following question, it should be understood that repition of a digit is not allowed.
(i) How many three-digit numbers can be formed from the five digits?

I was thinking we could do permutation for this. It would be $P(5,3)=60$

(ii) How many three-digit numbers which are less than 600 can be formed from the five digits?

I am not sure about this one. Can someone please show me?

(iii) How many three-digit numbers which are even numbers can be formed from the five digits?

I was thinking $4*3*2=24$ since $n_1 *n_2*n_3=n_k$
I was not sure of this problem since I just started learning this. I am not sure about (ii). What can I do for this part of the question? Can someone please tell me if this is correct? Is there another way of doing this?
 A: Your first is correct. 
For the second, for the leftmost digit, you need to look at which digits are less than $6$. All possible 3-digit numbers resulting when we choose the first digit to be less than six will be numbers less than $600$. There are exactly three such digits, so three options for the first digit. After that, any of the four remaining digits will work for the middle digit, and so on:
$$3\times 4\times 3$$
For the last, you are correct. We need for the rightmost digit to be even: there are two such choices, then there are $4, 3$ choices, respectively for the remaining two digits. That gives you, as you've shown: $4 \times 3\times 2 = 24$ such distinct three digit numbers.
A: Well for 2) you have the variants starting with 1,3 and 4, because a 3-digit number starting with 6 or 7 will be bigger than 600. For 3) by even I think you mean divisible by two, so now look at the last digit of the number it should be 4 or 6. I think you can take it from here. Also the first one is OK.
A: i) All the digits are distinct so the answer in $5 \cdot 4 \cdot 3 = 60$
ii) The first digit can be $1, 3, $ or $4$ so we have $3 \cdot 4 \cdot 3 = 36$ possibilities
iii) The last digit can be $4 $ or $6$ so we have $ 2 \cdot 4 \cdot 3 = 24$ possibilities
