Simple Combination problem My mathematical background in school involves no probability being taught, so this question might be very straightforward to answer. 
I have five close friends and we all want to have lunch together. My school has 3 lunches, A, B, and C. We know this probably won't happen, but I was wondering what the probability would be for exactly two, three, four or five of us to have lunch together.
For any two of us to have lunch together: The chance of me getting a specific lunch A, B, or C is 1/3. I can apply this logic to two people having a lunch together as 1/9. But I know that no matter the combinations of people in lunches it is more rare for 1 person to be alone at lunch than for 2 people to be alone, but I am very unsure about my thinking.
 A: In how many cases is each lunch being eaten by $2$ people each? (we represent this by $(2,2,2)$. Well, there are ${6 \choose 2} = 15$ ways to pick two people to eat lunch $A$, and then ${4 \choose 2}=6$ ways to pick two people to eat lunch $B$, leaving the others with lunch $C$. So that is $15*6=90$ possibilities.
As another example: how about 3 people eating one lunch, 2 other people eating a different lunch, and the last person eating the third type of lunch (this would be $(3,2,1)$? Well, let's first consider the case where the $3$ people eat lunch $A$, the next two eat lunch $B$, and the last person eats lunch $C$. There are ${6 \choose 3} = 20$ ways to pick three people to eat lunch $A$, and then ${3 \choose 2}=3$ ways to pick two people to eat lunch $B$, leaving the last person with lunch $C$. So that is $60$ possibilities. Of course, the same holds for $3$ people eating lunch $B$, 2 people eating lunch $C$, and $1$ person eating lunch $A$. Since there are $6$ different ways to change this, you get a total of $360$ possible ways for $3$ people eating one lunch, $2$ other people eating a different lunch, and the last person eating the third type of lunch.
Likewise, we can find the numbers for the different possibilities:
$(2,2,2)$: ${6 \choose 2} \cdot {4 \choose 2} = 15 \cdot 6 = 90$
$(3,2,1)$: ${6 \choose 3} \cdot {3 \choose 2} \cdot 6 = 360$ (the most likely outcome .. this happens almost half of the time)
$(3,3,0)$: ${6 \choose 3} \cdot 3 = 20 \cdot 3 = 60$
$(4,2,0)$: ${6 \choose 4} \cdot 6 = 15 * 6 = 90$
$(4,1,1)$: ${6 \choose 4} \cdot {2 \choose 1} \cdot 3 = 15 * 2 * 3 = 90$
$(5,1,0)$: ${6 \choose 5} \cdot 6 = 6 \cdot 6 = 36$
$(6,0,0)$: ${6 \choose 6} \cdot 3 =3$
Total: $729$ possibilities, which indeed equals $3^6$ *(sanity check!)
If you want the probabilities for these events, just divide the number of possible ways these events can take place by $729$. So, for example, $P(3,2,1)=\frac{360}{729} \approx 49.3$%
