Brain Teasers
Madadian Multiplex
Probability
Probability puzzles require you to weigh all the possibilities and pick the most likely outcome.Probability
The Madadian Multiplex Cinema is showing the latest offering from Arnold Schwartzandeggburger, a multi-sequel film called "Conan the Terminating Twin Recall Running Kindergarten Action Hero." A group of friends, consisting of eleven boys and girls, wait to take their seats in the same row in a movie theater. There are exactly 11 seats in the row.
They decided that after the first person sits down, the next person has to sit next to the first. The third sits next to one of the first two and so on until all eleven are seated. In other words, no person can take a seat that separates him/her from at least one other person.
How many different ways can this be accomplished? Note that the first person can choose any of the 11 seats.
They decided that after the first person sits down, the next person has to sit next to the first. The third sits next to one of the first two and so on until all eleven are seated. In other words, no person can take a seat that separates him/her from at least one other person.
How many different ways can this be accomplished? Note that the first person can choose any of the 11 seats.
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Gee...I wonder if he's trying to hint at some big movie star....? 
One easy way to see this is to imagine
that we are trying to assign kids to seats in reverse order so that the empty seats always form a contiguous block (this insures that the kids form a contiguous block when seating in the proper order).
When you look at the problem in
this reversed way, it should be clear that we have two choices for locating each child except #1 (whose position is forced). Each set of choices leads to a unique order. Hence there are 2^10 possible orders.
that we are trying to assign kids to seats in reverse order so that the empty seats always form a contiguous block (this insures that the kids form a contiguous block when seating in the proper order).
When you look at the problem in
this reversed way, it should be clear that we have two choices for locating each child except #1 (whose position is forced). Each set of choices leads to a unique order. Hence there are 2^10 possible orders.
Very interesting problem. I guessed 1000 without doing any real calculations, which was not bad for a guess. Good teaser.
by the last explanation ( 2(N-1) ) i figured it must be 20 for 11 ppl!!!
2(N-1) should read 2^(N-1). ie: 2 to the power of (N-1).
Good teaser.
I like your explanation, dewtell.
I like your explanation, dewtell.
It's kind of funny that I got an answer 8 orders of magnitude greater than the given answer...and I think mine has to be the correct answer.
As the problem states: "how many ways can this be accomplished?" with 'this' meaning how the friends take their seats. So I multiplied the 1024 orders that the seats can be taken by the 11! = 39,916,800 different orders the friends can take seats giving 40,874,803,200 ways to accomplish 'this'.
I was also wondering how many rows of 11 seats the theatre had since the selection of the row would seem to be an implied variable in the problem as well.
As the problem states: "how many ways can this be accomplished?" with 'this' meaning how the friends take their seats. So I multiplied the 1024 orders that the seats can be taken by the 11! = 39,916,800 different orders the friends can take seats giving 40,874,803,200 ways to accomplish 'this'.
I was also wondering how many rows of 11 seats the theatre had since the selection of the row would seem to be an implied variable in the problem as well.
The answer is wrong - by the given logic, say if the first person chooses a corner seat all have just one choice and at anytime if the corner seat is taken, everyone just has one choice
I think the solution here is 11! as it is not said that the order is fixed, ultimately there are 11! ways of arranging the kid, and this doesnt leave any seat in the middle
I think the solution here is 11! as it is not said that the order is fixed, ultimately there are 11! ways of arranging the kid, and this doesnt leave any seat in the middle
If the eleven people can end up in any order in the eleven seats, there are 11! (factorial 11) possible orders. This is because the first can be any of 11 people, the second any of 10 remaining, etc I.e. 11 x 10 x 9 x 8 x 7 x 6 ... x1 = 39916800 according to my calculator.
It is possible to set conditions about how they sit down that would prevent some options, which is I presume the idea here, but I am not clear why this would be the case here - I'm not sure I follow the instructions though. Perhaps clarification?
It is possible to set conditions about how they sit down that would prevent some options, which is I presume the idea here, but I am not clear why this would be the case here - I'm not sure I follow the instructions though. Perhaps clarification?
On re-reading I see the question is about the ways of sitting down, with the identity of any particular person being regarded as irrelevant. Sorry. The question could be clearer but it's my mistake.
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