jane1820
Well-Known Member
can someone explain this, i dont understand
-
Looking for HSC notes and resources? Check out our Notes & Resources page
Baulkham Hills 2020 trials 14.c (1 Viewer)
- Thread starter jane1820
- Start date
cheesynooby
Well-Known Member
say there's 4 different books (n), and there are 3 copies of each book (p).
then there's 4 x 3 books in total
and so you are arranging 12 books in a line (np)!, however to account for overcounting divide by 3! for each type of book. (so (p!)^n)
for part 2, each book can either be borrowed, with p possibilities for which book to be borrowed, or not borrowed (1 possibility) so each book there is (p+1) ways to select a book. this is multiplied together n times for each different book, however subtract 1 for the possibility that no books are borrowed
then there's 4 x 3 books in total
and so you are arranging 12 books in a line (np)!, however to account for overcounting divide by 3! for each type of book. (so (p!)^n)
for part 2, each book can either be borrowed, with p possibilities for which book to be borrowed, or not borrowed (1 possibility) so each book there is (p+1) ways to select a book. this is multiplied together n times for each different book, however subtract 1 for the possibility that no books are borrowed
jane1820
Well-Known Member
cheesynooby
Well-Known Member
i don't think so? it's just all the possible ways of borrowing different books, but subtract 1 to account for the fact that u can't borrow 0 books
like say there's 2 books, and each has 10 copies. then there's 11 ways of choosing the first book (+1 in case u don't want to borrow it) and 11 for the second book, so 11^2 in total. and then subtract 1 since u can't not borrow both