Help with Math. I can't figure it out?

Question

A porter leads eight guests to their hotel rooms, rooms I through 8. Unfortunately, the keys are unlabeled and the porter has mixed up their order. Using trial and error; what is the maximum number of attempts the porter must make before he opens all the doors?

Answer 1

1st guest 8 attempts

2nds guest 7

3rd guest 6

4th guest 5

5rd guest 4

6th guest 3 attempts

7ths guest 2 attempts

8th guest 1 attempts

36 tries total maximum.  If the porter was very unlucky.

After the porter opens the first door...he only has 7 left to try.  Each time he loses a door to try.  64 would be if he tried to open each door with each key.  That would not make any sense.

Wagamuff assumes that a successful attempt does not count.  If his theory was correct then it should be 7,6,5,4,3,2,1,0  Because each time he is not counting the opening as an attempt.

It is an attempt.  A successful one.  Therefore 36 is still correct

Answer 2

avatarr2003 is nearly right.

each time you open a door, you eliminate one key, because if you knew a key opened door 1, you wouldn't try to use it to open door 2!

BUT also, by the time you reach door 8, you only have 1 key left, so you do not need an attempt to open that door - you already know which key it is.

So 8+7+6+5+4+3+2+0 = 35

Hope that helped =)

Answer 3

64

8x8

Answer 4

64 8x8 liek the first dude. Its really confusing when you read it the first time LOL i had to read it like 5 times before getting what the person trying to ask LOL

Answer 5

Well  the maximum number of attempts would be based upon he assumption that every door he tried to open took every key he had. For example, if he tried each of the eight keys on the first door and was unfortunate enough to try them all,(i.e. 8 keys), before he got the right one, then the first door would take eight tries. Using the same logic, the second door would take 7 tries, the third door 6 tries, and so forth. If he were indeed that unlucky, then the maximum number of tries would be: 8,7,6,5,4,3,2, and 1. If you add these tries up you get 35.