Using the digits 1, 2, 3, 4, and 5 to make a 5 digit number, where digits cannot be repeated, how many have 4 before 2? I understand how to determine how many total possible 5-digit numbers with no repeating numbers you can find, I am just not sure how to figure out the last part
 A: By symmetry, having 4 before 2 and 2 before 4 are essentially the same case. Since there are a total of $5!=120$ combinations, there are $\frac{120}{2}=60$ numbers with 4 before 2.
A: There are 5 non repeatable digits there are $ 5! $ possible sequences of such digits.
If there was a sixth digit in a fixed place (the end or beginning) there would still be $ 5! $ possible sequences of the digits.
The requirement is that 4 comes before 2 in the sequence. I am assuming 'first' means that 4 is to the left of 2.
There are several positions these digits can be placed in:
$$ 4XXX2 $$
$$ 4XX2X $$
$$ 4X2XX $$
$$ 42XXX $$
$$ X4XX2 $$
$$ X4X2X $$
$$ X42XX $$
$$ XX4X2 $$
$$ XX42X $$
$$ XXX42 $$
There are 10 total sequences where 4 can be before 2 if each other digit is the same.
For each of the ten sequences if the X's were replaced with digits 1,3, and 5, there would be $3!$ different possible sequences for each.
The number of possible sequences of 1,2,3,4 and 5 where 4 is before 2 is 
$$ 10 * (3!) = 60 $$
