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Apeksha2101
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Three dwarves and three elves sit down in a row of six chairs. If no 09 Apr 2024, 02:11
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Bunuel
Three dwarves and three elves sit down in a row of six chairs. If no dwarf will sit next to another dwarf and no elf wil sit next to another elf, in how many different ways can the elves and dwarves sit?

A. 18
B. 36
C. 48
D. 72
E. 96


In order to meet the restriction dwarves and elves must sit either DEDEDE or EDEDED. There are 3!*3!=36 arrangements possible for each case (3! arrangements of dwarves and 3! arrangements of elves), so total ways to sit are 2*36=72.

Answer: D.
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Clearly it is DEDEDE or EDEDED, which means OR (+), this means 3! + 3!= 12 , can you please explain
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Bunuel
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Three dwarves and three elves sit down in a row of six chairs. If no 09 Apr 2024, 02:31
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Apeksha2101

Bunuel
Three dwarves and three elves sit down in a row of six chairs. If no dwarf will sit next to another dwarf and no elf wil sit next to another elf, in how many different ways can the elves and dwarves sit?

A. 18
B. 36
C. 48
D. 72
E. 96


In order to meet the restriction dwarves and elves must sit either DEDEDE or EDEDED. There are 3!*3!=36 arrangements possible for each case (3! arrangements of dwarves and 3! arrangements of elves), so total ways to sit are 2*36=72.

Answer: D.
­

Clearly it is DEDEDE or EDEDED, which means OR (+), this means 3! + 3!= 12 , can you please explain
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No. For each of the cases, DEDEDE or EDEDED, the number of arrangements is 3!*3! - 3! ways to arrange dwarves and 3! ways to arrange elves. Since there are two cases, DEDEDE or EDEDED, we multiply 3!*3! by 2.

Hope this helps.
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Three dwarves and three elves sit down in a row of six chairs. If no 16 Dec 2025, 04:12
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We cannot solve saying - first arrange dwafs which is 3! and then in remaining space 4c3 elfs. This can lead to a case where two dwafs are next to each other and
EDED_DE

hence this menthod is only used when only one group is not alternate. here both the groups are not alternate.
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