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Math Help - Combination Permutation question....a bit complicated.

  1. #1
    Nov 2012
    sri lanka

    Combination Permutation question....a bit complicated.

    So basically its about how many possible exam paper sets can be formed.

    Each set contains 5 papers.

    The order of the papers do not matter.

    Each paper contains 4 questions. They are of different types. Lets say types A,B,C,D

    Each question type can be selected from a bank of 5 questions for that type. That is there are 5 type A questions, 5 type B questions all in all there are 20 questions in the bank.

    In a particular set of 5 papers, if a question of tye A is used for one paper, it cannot be used for the other papers. (same applies for types B, C, D)
    The order of question types in a paper is always A,B,C,D.

    How many possible paper sets are there?


    My attempt:

    for each paper...

    for the question type A : 5C1 = 5
    for type B : 5C1 = 5

    so with C & D the ways to select paper 1 = 5x5x5x5 = 625.

    now there are 4 questions left in each type for constructing paper 2 = 4x4x4x4 = 256

    and similarly for paper 3 = 3x3x3x3 = 81

    and paper 4 = 2x2x2x2 = 16

    and paper 5 = 1x1x1x1 = 1

    so altogether the possible sets are = 625 x 256 x 81 x 16 x 1 = 207,360,000

    this is the permutations of the 5 paper sets. Since order of the 5 papers doesn't matter in each set, we divide by 5!

    so final answer is = 207,360,000/5! = 1,728,000

    Is this correct??? It seems a rather lot of combinations!!!


    There is also another part to this problem...

    find the answer if every paper in every set must have at least 2 questions different from every other paper.

    I have no idea how to tackle that. I think the previous answer gives it including only 1 question being different.
    Last edited by shadowvalar; November 7th 2012 at 08:41 AM.
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