<s>[INST] Choosing group of size 3 in a roll of 5 unique dice Suppose you have $5$ fair dice (with sides $1-6$) colored red, green, yellow, orange and blue.
In how many ways can you roll them such that the resulting set of numbers rolled is of size $3$? For instance, the case where we roll: $1,2,3,1,2$ is accepted but $1,2,3,4,1$ is not.
My attempt:
Choose the set of size $3$ to work with: ${6 \choose 3}$
Choose $3$ dice to insert those three unique numbers into: $5 \choose 3$
Assign the numbers in the set to colors (we have to use all three numbers): $3!$
In the two remaining dice we have three options: $3^2$
overall: ${6 \choose 3}\cdot{5 \choose 3}\cdot3!\cdot3^2$.
Is this correct?
 [/INST] You can either have three of one number and two singletons or two of two numbers and one singleton.  For the first, you have $6$ ways to choose the number there will be three of, $5 \choose 3$ ways to choose the dice with that number, $5$ ways to choose number on the first of the other dice and $4$ ways to choose number on the last, for $$6\cdot {5 \choose 3}\cdot 5 \cdot 4=1200$$
For two pairs plus one, you have $5 \choose 2$ ways to pick the first pair and $6$ ways to pick its number, $3 \choose 2$ ways to pick the second pair and $5$ ways to pick its number, then $4$ ways to pick the number for the last, but we have double counted because we can swap the two pairs.
$$\frac 12{5 \choose 2}6{3 \choose 2}5\cdot 4=1800$$
For a total of $3000$
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