Probability - ways in which its possible

[JessBrown]

So there are four books to be read by 2 people, same time to read a book each, and the question is how many different ways can they read the four books?

So I firstly took the approach that you have:
Person 1: Spot1 Spot2 Spot3 Spot4
Person 2: Spot1 Spot2 Spot3 Spot4
And then filling in how many book options there are in each spot.

So Person 1 Spot1 has 4 options.
Whereas Person2 Spot1 then has only 3 options (as one book is being read).
Therefore Person 1 Spot 2 then has 3 options left (Person1 has read one book already).
My biggest issue is in Person 2 Spot 2, I think there are 3 options, but not sure! (As Person2 has read one book, so has three remaining??) (or only 2 options?)

Then Person 1 Spot 3 has 2 opitons, Person 1 Spot4 has 1.
Person 2 Spot 3 I think has 2 remaining (or 1) and Person 2 Spot 4 has 1

Sooooo I think there are
Person 1: 4 3 2 1
Person 2: 3 3 2 1
Which results in 432 different ways in which to read the books... is this right though cos I have a feeling it may be
Person 1: 4 3 2 1
Person 2: 3 2 1 1
Which means there are only 144 options?

Thanks!

 

[bpet]

This one's a nice twist on the old derangements problem. The correct answer is somewhere between the two numbers you guessed and a bit trickier to compute.

Ok, so you found that person 1 can read the books in 24 different ways. Suppose person 1 chose any order and then relabel the books 1,2,3,4 in that order. Person 2 now must choose an ordering so that 1 isn't 1st, 2 isn't 2nd etc. Now there are only 24 combinations to check which can be done by brute-force, i.e. write them all out and find the valid ones (it will be an odd number).

 

[JessBrown]

Sooooo I wrote them all down... and the answer I got was still 216, as there are nine combinations for each of the first row's orders... so 24 ways to order the first row, means 24x9 ways to order in total... the first way was so much quicker though!

eg.
1234

2143
2341
2413
3142
3412
3421
4321
4312
4123

gives nine ways for the combo of 1234 on top row, therefore there are 24 different ways for the top row, 216 in total!

anyone want to check this for me? I am fairly sure now...

 

[bpet]

Well done! You can check the answer here: http://en.wikipedia.org/wiki/Derangement

 

[blue_raver22]

I would approach this problem using the fundamental principle of counting. This principle states that if one event can occur in m ways and a second event can occur in n ways, then the two events together can occur in m x n ways. In this case, the two events are the choice of books for Person 1 and Person 2.

So, for Person 1, there are 4 books to choose from for the first spot, 3 books for the second spot, 2 books for the third spot, and 1 book for the last spot. This gives us 4 x 3 x 2 x 1 = 24 options for Person 1.

For Person 2, there are also 4 books to choose from for the first spot, but then only 3 books for the second spot (since one book has already been chosen by Person 1), 2 books for the third spot, and 1 book for the last spot. This gives us 4 x 3 x 2 x 1 = 24 options for Person 2.

To find the total number of ways in which the books can be read, we multiply the options for Person 1 and Person 2 together: 24 x 24 = 576 possible ways.

So, the correct answer would be 576 different ways in which the four books can be read by two people at the same time. It is important to note that this assumes that each person can only read one book at a time and that the order in which they read the books does not matter. If these assumptions are not true, then the number of possible ways would be different.