Struggling in my Discrete math class, and working on this problem I've read the notes but i am lost on a few things. On the first part, I am loss between to use the combination formula of
$n!/(n-k)k!) $ as the description for this is n number k at a time. which would give me $360$
But I also heard of that just taking $5^{5^5}$ would give you the answer for (a)
Suppose the letters from the set $\{A, B, C,D, E\}$ are used to form strings of length 3.
(a) How many strings can be formed if we allow repetitions?
(b) How many strings can be formed if we do not allow repetitions?
(c) How many strings begin with A if we allow repetitions? (d) How many strings begin with A if we do not allow repetitions?
The total number of options is the product of the number of choices for each choice.
a) There's 5 choices for the first letter, for each of those five choices there are 5 choices for the second letter, and for each of those 25 choices of the first two letters, there are 5 choices for the third letter for a total of $5\times5\times5 = 5^3$.
b) There are 5 choices for the first letter. The second letter has to be different so there are 4 options for the second letter. The third letter has to be different so there are 3 options. This is a total of $5\times4\times3$ combinations.
c) There is only 1 choice for the first option. 5 for the second. 5 for the third. This is a total of $1\times5\times5$.
d) There is only 1 choice for the first. 4 (anything be A) for the second. And 3 for the third.
$n!$ is the number of ways to choice n items from n choices without repetition and where order maters. This is because you have $n$ choice for the first item, the second must be different so you have $n-1$ options for the second. $n-2$ for the third and so on. This is $n\times(n-1)\times(n-2)\times\ldots\times2\times1 = n!$ outcomes.
$\dfrac{n!}{(n-k)!} = n\times(n-1)\times\ldots(n - k +1)$. This is the number of ways to choice k items from n choices without repetition. (Same reason).
$\dfrac{n!}{(n-k)!k!}$. This is the number of ways to choice $k$ items from n choices without repetition when order doesn't matter. This is because there are $n!(n-k)!$ ways to choice $k$ items in order, and there are $k!$ ways to order these $k$ items. ($k$ choices for the first, $k-1$ for the second, …). So the total number of ways to choice $k$ items when order doesn't matter is
$$\frac{\text{Total number where order matters}}{\text{Total number of ways to order them}} = \frac{\frac{n!}{(n -k)!}}{k!} = \frac{n!}{(n-k)!k!}$$
So for in your example b)
How many ways are there to choice where order matters is: $$\frac{5!}{(5 - 3)!} = \frac{5!}{2!} = 5\times4\times3$$
If order doesn't matter the $ABC$ is the same as $BCA$ is the same as $CAB$ etc. for every ordered $XYZ$ choice there are $3\times2\times1 = 3!$ ways to arrange these choice so there are
$$\frac{5!}{(5 -3)!3!} = \frac{5\times4\times3}{3\times2\times1} = 10\,\text{options}$$
They are:
ABC (which is the same as BCA etc.)
ABD
ABE
ACD
ACE
ADE
BCD
BCE
BDE
CDE
Does that help?