By Zephyrmaster2 (Zephyrmaster2) on Saturday, October 11, 2003 - 09:27 pm: Edit

I didn't get it. I got 4.

1= car
2= no car

2111
1211
1121
1112

 By Culov (Culov) on Saturday, October 11, 2003 - 09:30 pm: Edit

if you want to do it visually, think about it as car x, car y, car z, and an empty spot.

That leaves many more possibilities than 4. Actually 24. 4(3!). For some reason i doubted my reasoning and changed my answer at the last second... I HATE IT WHEN I SECOND GUESS MYSELF!!!

 By Crypto86 (Crypto86) on Saturday, October 11, 2003 - 09:30 pm: Edit

You're logic is flawed. Their are 3 cars and 4 spaces. So for every permutation, there will be the 3 cars arranged in a way and the remaining space. Even if you just arrange the cars in all different ways you get 6:

123
132
213
231
312
321

Anyways, if you count the empty space as a "car", you will have 4 cars and 4 spots. And the permutation arrangement of n objects is n! (Or n times all the consecutive lower integers to 1, including 1)

So... 4! = 4*3*2*1 = 24

Culov - that sucks man. I hate when I do that too. Anyways, good pickup on that n! = n(n-1)! - So 4(3!) works too.

 By Zephyrmaster2 (Zephyrmaster2) on Saturday, October 11, 2003 - 09:31 pm: Edit

Oh... separate cars...

 By N350zs (N350zs) on Saturday, October 11, 2003 - 11:09 pm: Edit

use the formula (N!)/(N-r)!

= (4!)/(4-3)!=24

 By Sup4 (Sup4) on Sunday, October 12, 2003 - 11:43 am: Edit

simle way to solve it
first select 3 parking lots from four
4c3
=4!/3!
now arrange the cars in the three parking lot
= 3! ways

therefore ans = 4!/3!*3!
4!