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30 Apr 2016, 00:54
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D
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Difficulty:
85%
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Question Stats:
46% (01:51) correct
54%
(01:50)
wrong
based on 193
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Five friends go to a city where they can stay in 10 different hotels. In how many ways can they stay in these hotels if atleast two of them stay together?
(A) \(5^{10} - 10C5*5!\)
(B)\(5^{10} - 10C5\)
(C)\(5^{10} - 5!\)
(D) \(10^5 - 10C5*5!\)
(E) \(10^5 - 10C5\)
Self Made
(A) \(5^{10} - 10C5*5!\)
(B)\(5^{10} - 10C5\)
(C)\(5^{10} - 5!\)
(D) \(10^5 - 10C5*5!\)
(E) \(10^5 - 10C5\)
Self Made
goforgmat
Joined: 09 Feb 2015
Last visit: 02 Nov 2019
Posts: 235
Given Kudos: 232
Location: India
Concentration: Social Entrepreneurship, General Management
Schools: Booth '21 (D) Cambridge "21 (D) LBS '22 (D)
GMAT 1: 690 Q49 V34
GMAT 2: 720 Q49 V39
GPA: 2.8
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30 Apr 2016, 01:21
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b
Each has the option of staying in 10 different hotels ,total number of ways=5^10
This also includes combination where each stays in 5 different hotels=10c5
5^10 - 10c5
Each has the option of staying in 10 different hotels ,total number of ways=5^10
This also includes combination where each stays in 5 different hotels=10c5
5^10 - 10c5
30 Apr 2016, 01:24
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how many ways can they stay in these hotels if atleast two of them stay together --> Total - Number of ways so that no person stays together
There are 10 hotels and each person can stay in 10 hotels
Total = 10 * 10 * 10 * 10 * 10 --> 10^5 --> Notice that there is no restriction to avoid a hotel.
Number of ways so that no person stays together with another friend = 10 * 9 * 8 * 7 * 6 = 10P5 = 10C5 * 5!
Number of ways can 5 friends stay in these hotels if atleast two of them stay together = 10^5 - (10C5 * 5!)
Answer: D
There are 10 hotels and each person can stay in 10 hotels
Total = 10 * 10 * 10 * 10 * 10 --> 10^5 --> Notice that there is no restriction to avoid a hotel.
Number of ways so that no person stays together with another friend = 10 * 9 * 8 * 7 * 6 = 10P5 = 10C5 * 5!
Number of ways can 5 friends stay in these hotels if atleast two of them stay together = 10^5 - (10C5 * 5!)
Answer: D
