eleven boxes have been arranged in a row. eight rabits are to be placed in eight of them so that no two rabbits occupy the same box. find the number of ways this can be done if no two empty boxes are adjacent.

this is in the combination section of my handouts. can someone explain why it's a combination 'cause i orginally thought it was a permutation due to the word "arranged". or is it combination 'cause of the word "placed"?

i tried-- 11! (total ways to arrange boxes)- 9!2!
but this is using permutation right? so i don't think it's right...

Question

eleven boxes have been arranged in a row. eight rabits are to be placed in eight of them so that no two rabbits occupy the same box. find the number of ways this can be done if no two empty boxes are adjacent.

this is in the combination section of my handouts. can someone explain why it's a combination 'cause i orginally thought it was a permutation due to the word "arranged". or is it combination 'cause of the word "placed"?

i tried-- 11! (total ways to arrange boxes)- 9!2!
but this is using permutation right? so i don't think it's right...

anonymous · Answer

@ParthKohli ? @hartnn ? @UnkleRhaukus ? @dumbcow  @precal ? @radar  @Yahoo! ? help please! thank youuu!!

anonymous · Answer

btw i got 9 since i grouped the 2 empty boxes into 1 unit plus the 8 boxes with a rabbit. then you can arrange these 2 empty ones 2 ways.... but it's under the combination section and i'm not doing any combinations! :(

dumbcow · Answer

no it is a combination problem, the word "arranged" in the problem is referring to the fact that the boxes are placed in a single file row.
you are trying to figure out the number of ways to put 8 rabbits in 11 boxes without any empties next to each other

start with assumption that empty boxes Can be adjacent, then subtract the possible combinations where they are adjacent

so you start with 11 choose 8
then the case where 2 empties are next to each other:
 - how many places can the 3rd empty box go?
 - there are 9 available spots for each placement of the 2 empties (there are 10)
 - there are 9 ways to have 3 empties in a row
Notice: those 9 ways are being double counted
 -->    1  2  X   =  X  1  2  , both are counted separately even though its the same 3 adjacent boxes that are empty

 -- so the num of possible combinations where empties are adjacent is:
  9*10 - 9 = 90 -9 = 81

--> (11 choose 8) - 81  = 165 - 81 = 84

anonymous · Answer

@dumbcow the book says the answer is 8! C(9,3) :( i dunno whyy

dumbcow · Answer

its because i didn't factor in that the rabbits had to be ordered.
there are 8 different rabbits... there are 8! ways of arranging 8 rabbits

C(9,3) or 9 choose 3 is 84 or the num of combinations of having 3 empty boxes with no adjacents

anonymous · Answer

wow! right!! thank you! i understand it now! really appreciate it!!:)

dumbcow · Answer

yw