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the combinations formula would be for a slightly different question actually - out of 10 (n) people, how many ways are there to select 3 (r) people.
for your specific one, 3 people with a yes no choice then it'll be 8 possible combinations: 2^3
the way to build up those would be
assume 1 person2 possibilities
doesn't send a card - let's call that 0
sends a card - call that 1so 1, 0
then for the 2nd person, you do the same choice, but append the list of choices you had before
e.g.
00, 01, 10, 11then for the 3rd, same again - take the 4 from the list above, and duplicate, but with 0 before one set and 1 before the next
000, 001, 010, 011, 100, 101, 110, 111and so on. each time the possible choices double
4 people: 16 possibilities
5 people, 32 possibilities
6: 64
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duncs
Cheers.
Yes, it's just 3 options each yes or no.
Ie what are all the combinations of those three either sending or not sending me a card.
I found something referencing the n and r, but didn't understand how to apply it. n = names of people sending me cards, right?
But then how does r fit in?
When I write them in a table I get 18 entries, but I'd like to verify that, and also have more knowledge for the future if I come across a bigger number of options.