Combinatorics - Hexadecimals, how many possible numbers.... with ascending order? 
How many different numbers with 4 figures can you make out of the sixteen hexadecimal ’figures’ $\{0,1,2,3,4,5,6,7,8,9,A,B,C,D,E,F\}$ 

$16^4$ possible solutions

(i) How many numbers are ’real’ 4-figure numbers, meaning the first figure $ \neq 0$?

$16^4 - 16^3 = 61440$ numbers, because $0,16,16,16 $are the undesirable outcome

(ii) How many numbers from (i) have four different figures?

$15 \cdot 14 \cdot 13 \cdot 12 = 32760$ out of the $61440$ possible numbers

(iii) How many numbers from (ii) end with figure 0?

$15 \cdot 14 \cdot 13 \cdot 11$? 

(iv) How many numbers from (ii) have figures in increasing order?

This is where I am confused the most. I think it would be $1/4! \times 4$ since there are $4!$ ways to order $4$ numbers initiating with $1$ number, but there are $4$ numbers.
 A: 
How many $4$-character strings can be formed using the sixteen hexadecimal digits?

Your answer $16^4$ is correct?

How many $4$-digit numbers can be formed using the sixteen hexadecimal digits?

Your answer $16^4 - 16^3$ is correct.

How many $4$-digit hexadecimal numbers have distinct digits?

We have $15$ choices for the leading digit since we cannot use $0$, $15$ choices for the next digit since we can now use $0$ but cannot use the leading digit, $14$ choices for the third digit since we cannot use the first two digits, and $13$ choices for the fourth digit since we cannot use the first three digits.
$$15 \cdot 15 \cdot 14 \cdot 13$$

How many $4$-digit hexadecimal numbers with distinct digits end with $0$?

We have $1$ choice for the last digit, $15$ choices for the leading digit, $14$ choices for the second digit, and $13$ choices for the third digit.
$$15 \cdot 14 \cdot 13 \cdot 1$$

How many $4$-digit hexadecimal numbers with distinct digits have digits written in increasing order?

Observe that since the leading digit cannot be $0$ and the digits are increasing, the four digits of such a number must be selected from the set 
$$\{1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F\}$$ 
A $4$-digit hexadecimal number with strictly increasing digits is completely determined by which $4$ of these $15$ digits are selected since once the four digits are chosen, there is only one way to arrange them in increasing order.  Hence, there are 
$$\binom{15}{4}$$
such numbers.
A: Edit: Thanks to commenters.
Presumably what's being asked here is how many four digit hexadecimal numbers are there whose digits are all in increasing order. We also have the restriction that $0$ can't be chosen, because, given digits of increasing order, this would imply $0$ as the first digit, which isn't allowed for a number in the proper format. So, we must have $4$ increasing digits chosen from $1$ to $15$. Increasing implies the digits are distinct. Since any choice of $4$ digits gives rise to exactly one such number and vice versa, we just need to count the number of sets of $4$ distinct digits chosen from the set of $15$. This is simply $$15 \choose 4$$
A: i) is fine.
ii) There are $15$ possible values for the first digit, $15$ for the second digit, $14$ for the third and $13$ for the last digit. Giving $ 15 \times 15 \times 14 \times 13$.
iii) There are $15$ possible values for the first digit, $14$ for the second digit, $13$ for the third and $1$ for the last digit. Giving $ 15 \times 14 \times 13 \times 1$.
iv) The first digit cannot be zero so there are $ \binom {15}{4} $ ways to choose $4$ digits , now put them in ascending order. 
