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I've got a task about combinatorics and I'm very unsure how to handle it. The task says, that I have four groups of people:

  • Group $A$ with $24$ people
  • Group $B$ with $28$ people
  • Group $C$ with $17$ people
  • Group $D$ with $18$ people

I have to get the combinations I can get, when I choose $2$ out of each group, so that I have $8$ people at the end.

I know that I have to get the combinations of the two people of each group first. Which would be for $A$: $276$ combinations, $B$ gets $378$, $C$ gets $136$ and $D$ with $153$. But I'm very confused what to do with these. Am I summing them up and calculate the combinations of $8$ people I can get from these?

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    Say we have only two groups: Group A with 3 people and group B with 2 people, and suppose we pick only one person out of each group. What do you think the answer should be then? – 2017-01-15
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    Since A gives me 3 options and B 2, I'd multiply them. The result would be 6. In my example I would get 2,170,857,024 possible combinations, what was my first idea but it just sounds so ridiculous big. – 2017-01-15
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    It may be big, but it's also correct, since for each pair of people we choose from group A, we can independently choose a pair for group B, and independently of that for group C and D. Therefore the number of options multiply in each step, – 2017-01-15
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    Thank you a lot! That was weirdly easy to check it myself, so thanks for the kickstart. – 2017-01-15

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After evaluating the combinations of groups $A$ to $D$, apply the basic counting principle to find the total number of possible results. Therefore, we have $276\times 378\times 136 \times 153 = 2{,}170{,}857{,}024$.