A die is rolled 5 times. How many outcomes will result in 5 different numbers? How many outcomes will have the 5th number equal to an earlier number? 
A die is rolled $5$ times. How many outcomes will result in $5$ different numbers? How many outcomes will have the $5$th number equal to an earlier number?

For the first part I did:
$$6 \times 5 \times 4 \times 3 \times 2 \times 1$$
so that we have $5$ different numbers. 
For the second part, I did:
$$6 \times 5 \times 4 \times 3 \times 2 \times 4$$
where the last $4$ repeats one of the $4$ previous values. 
Is this right? 
 A: The numeric answer in the first is correct, but if you are rolling $5$ times the $\times 1$ should not be there.  
For the second, you certainly should not have the $\times 2$ in there.  Your answer would be correct if the question were how many ways to get four different numbers in the first four rolls, then match one of those four on the fifth roll.  I do not see the requirement for the first four to be distinct, so would think we should be counting $12332, 12333,$ and $11111$ as successes as well.  You need to add the number of ways to get three different numbers on the first four rolls times three and similarly for two and one.  
You can condition on the number of different numbers in the first four rolls.  There are $6 \times 5 \times 4 \times 3 \times 4$ ways to get four different numbers, then match one on the fifth roll.  There are ${4 \choose 2}\times 6 \times 5 \times 4 \times 3$ ways to get two numbers the same out of the first four, then match one of the first three with the fifth roll.  There are ${4 \choose 3} \times 6 \times 5 \times 2$ ways to get three of the same in the first four rolls and match a number with the fifth.  There are $\frac 12 {4 \choose 2}\times 6 \times 5 \times 2$ ways to get two pair and match one.  Finally there are $6$ ways to get five of the same.  If I punch the calculator right, I get $4026$ 
A: You could think as follows:
For the first part you have it right, for the first throw you have $6$ distinct possibilities, for the second throw you have $5$ and so on. Using the so called product rule you end up with 
$$
6\cdot 5\cdot 4\cdot 3\cdot 2
$$
distinct outcomes.
For the second part your reasoning is not correct. First of all it assumes that all the previous throws are distinct which is not required (as the question stands now) so you missed the outcome
$$
1\ \ 1\ \ 2 \ \ 1\ \ 2
$$
for example.
You could reason instead as follows
$$
\#\text{outcomes with last throw is not unique}=\#\text{all outcomes}-\#\text{outcome with last throw is unique}
$$
where $\#$ stands for "number of". 
All possible outcomes are easy to enumerate, I put it here but try to find it yourself before you peek (you can peek by mouseovering the yellow part below)

 $6^5$ since all $5$ throws can take any of the $6$ possible values

The number of outcomes where the last throw is uniqe is a bit trickier. You could argue that the last throw is going to be some value it actually can be all $6$ possibilities and this cannot be thrown before so you have one less possibility in each of the first $4$ throws

 $6\cdot 5^4$ since as we said you have one less possibility with each previous throw

Putting these together you arrive at

 $6^5-6\cdot 5^4$

Hope this helped
