#pigeon hole principle?
36 messages · Page 1 of 1 (latest)
i have a slight feeling its the pigeon hole principle but i dont know how to apply it
Yeah it is, the idea here is that your "pigeon holes" should be the different pairs you wrote up
So 10 11 is one pigeon hole, 4 17 is another
can you see how to go from here?
thankssss
wait so would it be like
i have to pull out 10 pairs or something
not quite
im so confused
You only need to get 1 pair right
yes
the second you have both cards in a pair you win
what's the minimum number of cards you need to pull out to ensure you have both cards in some pair?
Wait do i need to pull out 11 cards
would the 11th card be the partner of one of the 10 cards i chose
Yeah that's basically it, you have 10 different pairs, and even if the first 10 cards you pick belong to different pairs, the 11th then has to belong to the same pair as some other card
omg thank u i need to do more pigeon hole questions tbh
i just have one more question im having trouble w
whats wrong w my answer for C
its wrong but i dont get how to do it idk where to start
i just subtracted where all teachers r next to each other to the total amount of permutatioms
ur method doesnt countfor two teachers being next to each other
so i think you should count that as well
yes subtract these too
normally i just count the number of possible arrangements
ye this question can be done using 'gap method'
I think the trick is to
is to order it like this
{{1,19},{2,18},{3,17}...{9,11},{10, null}}
Hence if you have 10 pigeon holes, you are guaranteed to pick a number from each one of the subsets
hence if you have 9, then you are guaranteed to get one extra number in one of the pigeon holes
doesn't tip it over 21 though
