Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Help me please

  1. Apr 14, 2004 #1

    xt

    User Avatar

    I got this two problems, I can't figure them out...

    A bookshelf contains m different books and n copies of each. How many different selections can be made from them?

    and

    In how many different ways can four letters be posted in four envelopes so that no one receives the correct letter?
     
  2. jcsd
  3. Apr 14, 2004 #2
    for the first question maybe it is:
    m*(m+n)

    but i only checked it for two specific cases.


    edit:
    after checking again i think the answer should be:
    if m>n than S=m*(n+m)
    if m<n than S=n*(n+m)
     
    Last edited: Apr 15, 2004
  4. Apr 14, 2004 #3

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    If n=1, then the answer ought to be 2^m, so which examples did you check?
     
  5. Apr 14, 2004 #4

    pig

    User Avatar

    1. For each book, you can take 0 of that book, or 1 of that book, ..., or n of that book. That is (n+1) possibilities. Since we have m books, the formula is (n+1)(n+1)(n+1)...(n+1) m times, or (n+1)^m.

    One of the possibilities is that we select 0 of every book, so if "none of the books" is not a valid selection, then it is (n+1)^m - 1.
     
    Last edited: Apr 14, 2004
  6. Apr 15, 2004 #5
    i didnt try it for n=1.
    the cases are:
    m=2 n=3
    and m=2 n=4
    perhaps the flaw is i should have put in the second case m different than 2.
    bougar :frown:
     
  7. Apr 15, 2004 #6
    i tried with n=1 and m=2 and here are the combinations i got:
    (1,1)
    (1,0)
    (0,1)
    if you put it in your formula it equals to 2^2=4 which is one combination more than in reality.

    (unless you include (0,0) a selection which i havent included in my two examples).
     
  8. Apr 15, 2004 #7

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    The empty selection is always a selection.

    pig's answer is correct assuming that the wording implies that picking any one of the n copies is the same event as picking any of the other of the n copies.


    The second question follows from the inc-exc principle.

    Let V_n be the event that the nth person gets the right letter, U_n the even they don't

    Then doing the event you want is the intersection of the V_i which is the complement of the union of the U_i, which can be found using inclusion exclusion
     
    Last edited: Apr 15, 2004
  9. Apr 15, 2004 #8

    xt

    User Avatar

    whats the logics behind this? why do you assume its a selection first then after the solution is done minus it off and say its not a valid selection?

    what is that? where can i learn this stuff from? elementary probability?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Help me please
  1. Please help me (Replies: 1)

Loading...