If there is a parade with six floats in six certain positions, how many ways can those floats be arranged so that no float is in its original position?

Question

bloomlocke367 · Answer

do you have any ideas?

anonymous · Answer

imma just give a guess and say 12

anonymous · Answer

I think it is 6! ways = 6x5x4x3x2x1 = 720

anonymous · Answer

I know the total combinations available are 6! or 720. I've been spending hours trying to figure out a way to remove the unnecessary combinations. For example any combination beginning with 1 in the first position is wrong, so that gets rid of 720/6=120 combinations, but I've had a lot of trouble after that.

anonymous · Answer

I'm not sure if I'm right.

Try using the indirect method.

#of ways float can not go back their original position: #total number of ways they can be arranged - #number of ways they can go back to their original position

Now you just have to find the #number of ways they can go back to their original position, which I think is only one 1 way because they each has only one original position. What do you think?

anonymous · Answer

So, if none cannot be in the same position,

There are 6C6 arrangements where all 6 of them are in the same position. So that's 1.

There are 6C5 arrangements where 5 of them are in the same position. So, that's 6.

There are 6C4*2! arrangements where 4 of them are in the same position. So, that's 30.

There are 6C3*3! arrangements where 3 of them are in the same position. So that's 120.

There are 6C2 arrangements where 2 of them are in the same position. So that's 360.

There are 6C1 arrangements where 1 of them are in the same position. So that's 120.

Adding them up, 637 arrangements where at least 1 of them is in the same position.

If my interpretation is correct, there shall be no floats in the original position. Where as, having 1 of them in the same position will not be counted as one way.

So there are 720 possible ways, 720 - 637 = 83 ways that there are NO floats in the same position.

Though I'm not that sure. Haha :)

anonymous · Answer

Data, I've been trying that for quite a while. It's nearly impossible to manually count all cases to see which ones return to their original position, even if I can eliminate obvious cases that won't work.

anonymous · Answer

@killua_vongoladecimo  I really think you're onto something here, although I don't have the answer myself. Should I be using combinations or permutations? I feel as if the latter would work better as order matters in this case. Can I also ask why you multiply the 4 and 3 choices by 2! and 3!?

anonymous · Answer

Actually, I used combinations. Because there are some of which, constant which kind of means, they did not change positions. 

Wait let me recall

anonymous · Answer

There are 6C4*2! arrangements where 4 of them are in the same position. So, that's 30.

In here, there are 6C4 ways in choosing 4 which will be in the same position. So if that happens, there are 2 slots left. So the number of ways the other 2 which moved is 2!

anonymous · Answer

@killua_vongoladecimo Also, if 5 of the numbers for the 6C5 case are in the original position, doesn't that mean the last digit is also in its original position?

mathstudent55 · Answer

This can be done with a simple program.