# Combination Permutation question....a bit complicated.

• Nov 7th 2012, 06:57 AM
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 etc....so 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)
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The order of question types in a paper is always A,B,C,D.

How many possible paper sets are there?

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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!!!

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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.