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How would you calculate all possible combinations of groups size 2 or 3 from 11 people

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If you just have to choose then it will be 11C2 and 11C3 respectively . Had you asked for arrangement of these then it would have been 11P2 and 11P3 , because this time 3 choses objects can be arranged in 6 ways also. So you would have to multiply 11C3 * 3! Which is 11P3.

You can think of it in this way -

If you just need to choose then your first option has 11 choices. For each choice you have another 10 options and for each 10 you have another 9 . But since there are no arrangements you got to divide be 3! since you would be over counting the different arrangement of the 3 persons.

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    So it would be 495? – 2017-02-20
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    @trebor97351 . No 11C2 is 55 – 2017-02-20
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    So the answer to the question: 11 people go to a quiz, were the maximum group size is 3, nobody wants to go on their own. In how many different groupings can they enter the quiz? – 2017-02-20
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    @trebor97351 If you have to chooses 3 out 11 in a random way then it's 11C3 =165 – 2017-02-20
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With binomials, so $\binom{11}{2}$ or $\binom{11}{3}$ respectively, where in general a subset of $k$ from a group of $n$ is $\binom{n}{k}=\frac{n!}{k!(n-k)!}$

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    Will that give all combinations, e.g 2,2,2,2,3? – 2017-02-20
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    @trebor97351 what is 2,2,2,2,3 supposed to be an example of? You said groups of size 2 or 3. Also combinations usually means without repeated objects. Do you want to include repeats? – 2017-02-20
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You can use the formula to do this. But here is the concept behind it.

Suppose you want to choose 3 people from 11 people.

In order to do that, imagine that there are 3 chairs. There are 11 ways to fill chair 1. After filling chair one there are 10 ways to fill chair 2 and similarly, there are 9 ways to fill chair 3.

Say chose 3 people A, B and C. Choosing any 3 people doesn't depend on what order we choose them in.

ABC, ACB, BAC, BCA CAB, and CBA are all the possible permutations of ABC but they all have the same people. Permutation is an ordered combination. Therefore, we can say that 6 permutations are actually 1 combination.

Thus, $Total\space Combinations =\frac{11.10.9}{3.2} $

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    Does that only calculate the first set of 3, or all different sets? – 2017-02-20
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    All the possible combos that can be made considering all the people...But each combo contains only 3 people... – 2017-02-20