how many ways can you place 20 different books on5 shelve sothere is at least two bookds on each shelve?
926536
One way to attack this is to see how many ways 20 books can be put  in a row and then see how may ways that row can be divided into 5 shelves. \n\nFor the row of 20, there are 20 choices for the first book, for each of the 20 choices of the first book there are always 19 left for the second giving 20*19 ways of picking the first two books.  Extending this on there are 18 choices for the third, etc.  So thee are 20*19*18*17.......*4*3*2*1or 20! ways to arrange the 20 books in order. That's 2.43 x 10^18.\n\nNow for each ordering, there are different ways to divide the order, without shuffling, among the shelves.  Since we have the books in order, we will keep them that way with the first set going in order to shelf one, next group to two, etc.  That way, we only need to figure how many ways the 20 books can be divided without regard to which book is which.  Each ordering above can go into any of the schemes to divide the books among the shelves so you just multiply the two numbers together to get the total number of possibilities.\n\nSince each shelf must have two books minimum, start by putting two books on each shelf and leaving them there.  The problem then comes down to how many ways can the remaining 10 books be divided among 5 shelves.  We really want to put 4 divisions along the row of 10 books to tell where one shelf ends and the other begins.  This gives 14 locations to be occupied by either a book or divider.  The number of possible arrangements of the 14 locations is 14!.  But since all dividers look the same and all books (for now) look the same, there are only 14!/(10! * 4!) = 1001 possible ways to divide it onto the shelves.\n\nSo there are 1001 ways to divide the 20 books among the shelves for each of the 20! Arrangements of books.  Multiplying these together gives 1001*20! = 2.43 x 10^21 ways to arrange the books.  If you want to see all the arrangements, I suggest you start arranging books right away.  If you put together one arrangement of books per second it would take about 5000 times the age of the universe to go through all of them.
289381
The answer to the three switch problem is 8 choices, not 7. The answer figured numerically is 2^3.\n\nWith 8 switches there are 2^8 or 256 combinations possible.\n\nThere is a formula based on the length of the femur:\nhttp://ublib.buffalo.edu/libraries/projects/cases/bones/activity4.html
18441
A library has over 1000, so I think you need 1000, sometimes with 200 you have already seen them all.  You need to have ones you have never seen, just like libraries have books you have never read.
236101
permutation?\n\ni'm not sure what you mean by "see how many it is."
812841
Given 5 children there are 2^5 or 32 possible combinations of girls and boys.  Since there is only one combination with all boys that leaves 31 combinations containing at least 1 girl.  Therefore the probability is 31/32.
366888
I cannot help you with the name of any specific book, because I do not know what your students might like; but I can help you finding the 30 copies; there are several sites where you can read the books for free and I think it would be your best and cheapest option.\nhttp://www.thefreelibrary.com/\nhttp://www.promo.net/pg/\nhttp://onlinebooks.library.upenn.edu/
54351
2 people, 2 pages, in 5 minutes\n\nSo...\nIn twice the time, they can type twice the number of pages:\n2 people can type 4 pages in 10 minutes\n\nTo type 5x as many pages in the same amount of time, you need 5x as many people:\n10 people can type 20 pages in 10 minutes
631136
Total 20 nines (2 in 99) are there in 0 to 100.\n\n9, 19, 29, 39, 49, 59, 69, 79, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99
1127065
4 pages in 5 minutes\n\n1 page in 5/4 min.=\n\n1 page in  1.25 minutes\n\n300 pages (1 book)in= 1.25x300 = min.\n                    =375 minutes\n\n1 day 6 hours means 360 minutes\n\nin 10 days  360x10 = 3600 minutes\n\n3600 minutes and typing at the rate of 375 minutes per book\n\n3600/375 =9.6 books
527277
Catagorize your shelves by subject.  Then when you put the books on the shelf, if this is your personal library, I would put them in alphabetical order by title.  Then make a master list as you are setting it up, so that you always know what you have and where you put it.
586719
I know 67 positions... I have a word doc, I can send u if u want it ;)
673673
Use the multinomial coefficient, which is an extension of the binomial coefficient.\n\n10!/(2!2!2!2!2!) = 113,400\n\nBasically it works like this: if your have people A, B, C, ..., J, then there are 10! ways to arrange them.  One arrangement of these 10 people with 2 people in five groups looks like this\n\nAB|CD|EF|GH|IJ\n\nwhere the bars are separating the groups.  Now another arrangement of these people looks like this\n\nBA|CD|EF|GH|IJ\n\nbut everyone is still in the same group.  So is this one\n\nBA|DC|EF|GH|IJ\n\nand this one\n\nBA|DC|FE|GH|IJ\n\nand so on.  It turns out there are 2!2!2!2!2! = 32 ways to rearrange this without changing the groups, since each pair has 2! ways of being arranged.\n\nSo since there will be fewer ways to make 5 groups of 2 than there are ways to arrange the 10 people, you come up with \n10!/2!2!2!2!2! ways to put 10 people into 5 groups of 2.\n\n10C2 and 10C5 would not be correct.  10C5 is the number of ways to select 2 groups of 5, not 5 groups with 2.  10C2 is the number of ways to select one group with 2 people and one group with 8 people.
660119
Different texts will vary somewhat but all are close to the same range. 5,000-12,000.
674638
yes.\nrefer +2 c books for basics.
182775
there are 6 place for 15 student so:-\n\n15*14*13*12*11*10=answer\n\nor simply use combination since the arrangement is not important\n\n15C6
939247
No, but you can work it out on paper. I know because I did it once. You have five letters, so there should be 25 combinations, Of course, as two letters are the same that would make it 20. I think. Just give it a go. It's always more fun to work these tings out yourself.
1013887
You should know that when you look at bookshelves you have to know how wide the book is that you are wanting to replicate over and over again, the length of the book, so that you know how wide the bookshelf has to be, how high off the ground you want it, how many books you want to have in it.  This question is not a simple 100 books and a website, it's much more complicated and you haven't specified the criteria that is needed.