Click here to go to the

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 |

read a lot about this question

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!

Posting is currently disabled in this topic. Contact your discussion moderator for more information. |

Administrator's Control Panel -- Board Moderators Only Administer Page |