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Four travelers arrive in a town with five hotels. In how

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Senior Manager
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Four travelers arrive in a town with five hotels. In how [#permalink] New post 08 Aug 2007, 11:44
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1. Four travelers arrive in a town with five hotels. In how many ways can they take up quaters?

(A) 1024
(B) 625
(C) 512
(D) 120
(E) 5

2. Four travelers arrive in a town with five hotels. In how many ways can they take up quaters with each traveler in a different hotel?

(A) 1024
(B) 625
(C) 512
(D) 120
(E) 5

Please, additionally, explain the difference in your approach for solving each of these questions.
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Re: PS questions (four travelers) [#permalink] New post 08 Aug 2007, 12:20
Vlad77 wrote:
1. Four travelers arrive in a town with five hotels. In how many ways can they take up quaters?

(A) 1024
(B) 625
(C) 512
(D) 120
(E) 5

2. Four travelers arrive in a town with five hotels. In how many ways can they take up quaters with each traveler in a different hotel?

(A) 1024
(B) 625
(C) 512
(D) 120
(E) 5

Please, additionally, explain the difference in your approach for solving each of these questions.



1. Each of the four travellers can choose any of the five hotels, because there are no restrictions. hence the total possible scenarios are = 5^4 = 625

2. First traveller has 5 choices, second has 4 becasue he cannot choose the one chosen by first, third has 3 choices and 4th has 2.

hence, total no of ways = 5*4*3*2 = 120
Senior Manager
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Re: PS questions (four travelers) [#permalink] New post 09 Aug 2007, 07:07
pandeyrav wrote:
Vlad77 wrote:
1. Four travelers arrive in a town with five hotels. In how many ways can they take up quaters?

(A) 1024
(B) 625
(C) 512
(D) 120
(E) 5

2. Four travelers arrive in a town with five hotels. In how many ways can they take up quaters with each traveler in a different hotel?

(A) 1024
(B) 625
(C) 512
(D) 120
(E) 5

Please, additionally, explain the difference in your approach for solving each of these questions.



1. Each of the four travellers can choose any of the five hotels, because there are no restrictions. hence the total possible scenarios are = 5^4 = 625

2. First traveller has 5 choices, second has 4 becasue he cannot choose the one chosen by first, third has 3 choices and 4th has 2.

hence, total no of ways = 5*4*3*2 = 120


Can you explain a little bit deeper the answer for the first question? I always confuse rising to power with permutations and combinations, so I want to understand the unerlaying mechanism when to use power and when to use the other stuff.
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 [#permalink] New post 09 Aug 2007, 07:28
Quote:
1. Four travelers arrive in a town with five hotels. In how many ways can they take up quaters?

(A) 1024
(B) 625
(C) 512
(D) 120
(E) 5



Since there are no restrictions on how/where the travelers can sleep each traveler has their choice of the 5 hotels.

5x5x5x5 = 5^4 = 625

This is the maximum number of combinations without restrictions. So they could all sleep in seperate hotels or they could all sleep in the same on, doesn't matter. Each traveler has 5 choices.
  [#permalink] 09 Aug 2007, 07:28
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