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Combinatorics

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Combinatorics [#permalink] New post 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?
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CEO
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Re: Combinatorics [#permalink] New post 04 Aug 2009, 12:46
Expert's post
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
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Re: Combinatorics [#permalink] New post 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!
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Re: Combinatorics [#permalink] New post 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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Re: Combinatorics [#permalink] New post 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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