Roll 6 dice, find the number of outcomes with 3 distinct numbers. Suppose you roll six dice, how many outcomes are there with 3 distinct numbers.
My attempt:
First there are ${6 \choose 3}$ ways to choose these 3 distinct numbers.
We consider 3 cases;
Case 1: 3 pars of repeated numbers e.g. $223344$. There are ${3\choose 3}$ choices for the values, and for the ordering there are ${6\choose 2} \times {4\choose 2}$.
Case 2: One number reapeated 3 times, one number repeated twice, one number appears once. There are ${3\choose 1}$ choices for the thrice repeated number and ${2\choose 1}$ choices for the twice repeated number. Also there are ${6\choose 3}\times {3\choose 2}$ orderings.
Case 3: One number repeated 4 times and 2 numbers appearing once. There are again ${3\choose 1}$ choices for the number appearing $4$ times and the number of orderings is ${6\choose 4}\times2\times 1$.
Hence in total there are $${6\choose 3}\bigg({6\choose 2} \times {4\choose 2}+3\times2\times {6\choose 3}\times {3\choose 2}+2\times 1\times{6\choose 4}\bigg)$$
Is this correct? I am most uncertain about case 2, is it correct to choose the value for and then still order it?

This is correct, except you forgot to include the factor $\binom31$ for case $3$ in the final result. We can double-check the result using inclusion-exclusion: There are $\binom63$ ways to choose the $3$ numbers, and by inclusion-exclusion there are
$$\sum_{k=0}^3(-1)^k\binom3k(3-k)^6=3^6-3\cdot2^6+3\cdot1^6=540$$
ways to form strings of $6$ that use all $3$ of them, for a total of $10800$, in agreement with your result.
