-
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.
hugo7-
edit: got everything wrong, double-counted, and agree with duncs below that it's choices (2) to the power of people (3), so 8.
https://www.mathsisfun.com/combinatorics/combinations-permutations-calculator.html
-
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
hvsds
duncs
combinations and permutations pretty much covers it
If you wanted to check how many combinations there were when, say, 3 from 10 sent you a card
then from a 'n, choose r' type calc then n=10, r=3 and the number of combinations would be 120
https://www.calculatorsoup.com/calculators/discretemathematics/combinations.php
in your example, if you just have two choices then you'd probably most often find it illustrated as a binary tree 3 levels deep