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08 Aug 2007, 12:44
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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.
(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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08 Aug 2007, 13:20
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Vlad77
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
09 Aug 2007, 08:07
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pandeyrav
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.
