Combinatorics - throwing colored dice We throw 5 colored dice in the colors: blue, red, yellow, green, orange.
How many results are they for:
1) In total
2) At least one die has the number '3'
3) Exactly one die has the number '2' and exactly one die has the number '4'
4) The set of results contains only 3 members - what it means is that only 3 numbers will be in the results (e.g 1 1 1 2 3)
What I did is that:
1) In total there are $6^5$
2) I'll take the total number of combinations and subtract all the possible ways '3' is not picked. 
$6^5-5^5$
3) I'll choose one die to represent '2', and one to represent '4'
${4 \choose 2} \cdot 4^5$
I'm having trouble solving the last question, I'm also not sure about my solutions.
Can anyone please verify it and/or give me a suggestion for the last exercise?
Thanks a lot
 A: Your answers for $(1)$ and $(2)$ are correct.
(3) $5\cdot4$ ways to place the $2$ specified #s and $4^3$ ways to fill the other places, so $5\cdot4\cdot4^3$
(4) Patterns with only $3$ numbers are:
$3-1-1-0-0-0:$
$\left[\binom61\binom52\;\text{ways to choose the numbers}\right]\times\left[\frac{5!}{3!}\text{ ways to permute them}\right]$  
$2-2-1-0-0-0:$
$\left[\binom62\binom41\;\text{ways to choose the numbers}\right]\times\left[\frac{5!}{2!2!} \text{ ways to permute them}\right]$
Add up.   
A: 3) Is not right: because 2 and 4 are different numbers you have $\binom{5}{2}\cdot 2!=(5)_2=20$ ways to choose pairs of dice on 5 dice, one 2 and other 4 (this is named variations without repetitions). The others 3 dice have 4 values to take so the total will be $20\cdot4^3$.
4) My interpretation of this is that in a throw they are only 3 different values present. The different ways to have only 3 values are these groups ABCCC and ABBCC. 


*

*The combinations of groups of A, B and C members are $\binom{A+B+C}{A,B,C}$, this is the multinomial coefficient. For the first case is $\frac{5!}{3!}=20$, and for the second is $\frac{5!}{4}=30$. But for every ABC combination there are 2 indistinguishable groups, i.e., two groups of the same cardinality, so the total amount of distinguishable combinations when values would be placed are $(30+20)\frac12=25$.

*The ordered combinations of 3 different values over 6 possible are $(6)_3=120$. We need ordered combinations (=variations) instead of just combinations because we are putting the 3 different values on the groups A, B and C in any different order.
Then the answer for the four question is $25\cdot 120=3000$ distinguishable throws with 3 different values.
A: Your answers for 1) and 2) are correct.
3) You have 5 choices for the dice that will show '2', then 4 choices for the one that will show '4'. Others cannot be '2' or '4'. So there are $5.4.4^3$ possibilities
Question 4) is not clear
