Combinatorics : PS Archive
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# Combinatorics

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Intern
Joined: 13 May 2009
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04 Aug 2009, 12:38
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Three dwarves and three elves sit down in a row of six chairs. If no dwarf will sit next to other dwarf and no elf will sit to another elf, in how many different ways can the elves and dwarves sit?
CEO
Joined: 17 Nov 2007
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Concentration: Entrepreneurship, Other
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04 Aug 2009, 12:46
1) There is 2 possible options how dwarves and elves could be placed: dedede or ededed

2) There is $$P_3^3$$ options for 3 dwarves/elves for choosing 3 chairs

3) Finally, $$2*P_3^3*P_3^3 = 2*3!*3! = 72$$
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Intern
Joined: 13 May 2009
Posts: 18
Followers: 1

Kudos [?]: 40 [0], given: 1

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04 Aug 2009, 13:00
Two Options :-

1.) DEDEDE

2.) EDEDED

Persons Choices Seat Assigned

Dwarf A 6 Choices(1,2,3,4,5,6) #1
Dwarf B 2 Choices(3,5) #3
Dwarf C 1 Choice(5) #5
ELF A 3 Choices(2,4,6) #2
ELF B 2 Choices(4,6) #4
ELF C 1 Choice(6) #6

Final product 6*2*1*3*2*1 =72

If the same process is followed for option 2 we will get the same result of 72. In that case the answer should be 72+72?? Correct me if i am wrong!
Senior Manager
Joined: 17 Jul 2009
Posts: 299
Concentration: Nonprofit, Strategy
GPA: 3.42
WE: Engineering (Computer Hardware)
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04 Aug 2009, 14:01
lbsgmat wrote:
Two Options :-

1.) DEDEDE

2.) EDEDED

Persons Choices Seat Assigned

Dwarf A 6 Choices(1,2,3,4,5,6) #1
Dwarf B 2 Choices(3,5) #3
Dwarf C 1 Choice(5) #5
ELF A 3 Choices(2,4,6) #2
ELF B 2 Choices(4,6) #4
ELF C 1 Choice(6) #6

Final product 6*2*1*3*2*1 =72

If the same process is followed for option 2 we will get the same result of 72. In that case the answer should be 72+72?? Correct me if i am wrong!

because your Dwarf A could choose among all 6 seats, so that's including both option 1 and 2 there
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06 Aug 2009, 08:41
dedede or ededed

Hence 3! x 3! x 2 = 72
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Re: Combinatorics   [#permalink] 06 Aug 2009, 08:41
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