In this case, it’s not getting the answer which is important as much as how you get it. So, don’t Google unless you’re desperate.
This involves spatial reasoning.
In 3-dimensional space…
1 infinite plane cuts space into a max of 2 compartments.
2 infinite planes cut space into a max of 4 compartments
3 infinite planes cut space into a max of 8 compartments.
I was told it’s not 2^5.
How many for 5 infinite planes ?
Any analytical ways to solve this ? (Not visualized geometry)
I don’t know. But I can guess why it’s not progressive (like powers of 2): each of the first three divisions take advantage of an additional, previously ‘unused’ dimension (x axis, y axis, z axis; or visualize it as a plane lying horizontally, then another going vertically left to right, then a third one going vertically directly away from you).
But we only have 3 dimensions (sorry SF fans, but ‘time’ doesn’t count as the fourth dimension), so you can’t continue to cut every compartment into half with each additional plane. Some will just have to wait their turn until a plane can come to their area.
Having said all that, I’ll take a wild guess: I know you can add 4 more compartments with the 4th plane (picture one exactly parallel to any one of the previous 3), and I think that you can only add 4 more compartments with the 5th plane, so I’ll say 16 compartments is your total.
According to the bizarrely arranged post it notes on my desk, if you put the fourth plane at an angle not parallel to one of the previous three (for example, if the first plane is x=0, the second y=0, the third is z=0, and the fourth is x+y+z=4) you’re going to get 14 infinite compartments and one finite compartment.
The question for someone much better at this than I is where do you put the 5th plane (or where do you put all of them if the arrangement of the first four is not the best arrangement).
The reason I’m asking is because some people can count it in their head. One notable being Robert McNamara, former Sec. of Defense.
Thing is, I can visualize it with what I believe is the optimal intersection of the planes. But I can’t hold it all the way. I tend to “lose” it around after the third quadrant (270 deg). One of these days…
I think the 5th one adds 6 compartments (make it perpendicular to the 4th), which would give you 18 compartments.
Although like amarinth said, 4 planes give 15 compartments.
And I’ve visualized that before. Upto 4 is easy. It’s 5 which is the problem.
I say 26, and here’s why. Consider first the two-dimensional case. 0 lines gives 1 region. 1 line gives 2. 2 gives 4. 3 gives 7. 4 gives 11. Each time you add the number of lines that are already there, plus 1. This is because an arbitrary new line (ie, one that is not paralell to any of the others, or intersects any of the others at a pre-existing intersection) will intersect each of the others once. It’s a little hard to explain, but if a line intersects n other lines, it will be divided into n+1 parts, and it will make n+1 new regions. Okay, so the total number of 2-D regions for n lines is:
f(n) = 1 + sum(p, p=1…n)
= n(n+1)/2 + 1
Now the 3-D case is analagous. If you have a plane, and m planes intersect it, then that plane will have m lines drawn on it, and it will be divided into f(m) regions. That means that it will add f(m) new regions. So the total number of 3-D regions for n planes is:
g(n) = 1 + sum(f(p-1), p=1…n)
= n(n[sup]2[/sup] + 5)/6 + 1
g(5) = 26
And we don’t have to stop there. The solution for the analagous 4-D case would be:
h(n) = 1 + sum(g(p-1), p=1…n)
= (n[sup]2[/sup] - n + 12)(n + 1)(n + 2) / 24
Rereading my post, I think I come across as considering myself pretty authoritative. Please realize that I don’t have any sort of formal proof. That’s just my best guess. 
I hope Achernar’s answer (or at least the reasoning) is correct. I got asked this exact question at a job interview last week and that’s the answer I came up with.
And that’s good enough. It’s correct.
So, you have the analytical method.
Now, getting the patience and visualizing the damn thing.