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In how many ways can you sit 8 people on a bench if 3 of them must sit together? a) 720 b) 2,160 c) 2,400 d) 4,320 e) 40,320
i'm getting 'e' but that is not OA
Say 8 people are {A, B, C, D, E, F, G, H} and A, B and C must sit together. Consider them as one unit {ABC}, so we'll have total of 6 units: {ABC}, {D}, {E}, {F}, {G}, {H}, which can be arranged in 6! ways. Now, A, B and C within their unit can be arranged in 3! ways, which gives total of 6!*3!=4,320 different arrangements.
Re: In how many ways can you sit 8 people on a bench if 3 of
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12 Mar 2012, 09:30
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Gavan wrote:
thanks!
did you consider following arrangements? 1: -{D},{ABC}, {E}, {F}, {G}, {H}, 2: -{D},{E}, {F},{ABC}, {G}, {H}, 3: - ..
Please clarify.
6 distinct object can be arranged in 6! different ways, so 6 units {ABC}, {D}, {E}, {F}, {G}, {H} can be arranged in 6! different ways, which takes cares of all possible cases.
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29 May 2016, 22:57
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Gavan wrote:
In how many ways can you sit 8 people on a bench if 3 of them must sit together?
A. 720 B. 2,160 C. 2,400 D. 4,320 E. 40,320
In such questions, always tie the person that have to sit together. So we have effectively 5+"1" = 6 "persons" to arrange. They can be arranged in 6! ways. Now the 3 persons can themselves be arranged in 3! ways.
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24 Oct 2017, 07:35
Bunuel wrote:
Gavan wrote:
In how many ways can you sit 8 people on a bench if 3 of them must sit together? a) 720 b) 2,160 c) 2,400 d) 4,320 e) 40,320
i'm getting 'e' but that is not OA
Say 8 people are {A, B, C, D, E, F, G, H} and A, B and C must sit together. Consider them as one unit {ABC}, so we'll have total of 6 units: {ABC}, {D}, {E}, {F}, {G}, {H}, which can be arranged in 6! ways. Now, A, B and C within their unit can be arranged in 3! ways, which gives total of 6!*3!=4,320 different arrangements.
Answer: D.
Hope it's clear.
quick question regarding this: Since the question doesn't specify which 3 people need to sit together, how come we don't have find ways of choosing 3 ppl out of the 8 to act as unit? Like {DEF}, {AGF} etc etc.. I was thinking we would have to do 8C3 x 3! x 6!... can you tell me where I am going wrong with my logic? Thanks for your help!
In how many ways can you sit 8 people on a bench if 3 of
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29 Oct 2017, 11:20
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Ace800 wrote:
Bunuel wrote:
Gavan wrote:
In how many ways can you sit 8 people on a bench if 3 of them must sit together? a) 720 b) 2,160 c) 2,400 d) 4,320 e) 40,320
i'm getting 'e' but that is not OA
Say 8 people are {A, B, C, D, E, F, G, H} and A, B and C must sit together. Consider them as one unit {ABC}, so we'll have total of 6 units: {ABC}, {D}, {E}, {F}, {G}, {H}, which can be arranged in 6! ways. Now, A, B and C within their unit can be arranged in 3! ways, which gives total of 6!*3!=4,320 different arrangements.
Answer: D.
Hope it's clear.
quick question regarding this: Since the question doesn't specify which 3 people need to sit together, how come we don't have find ways of choosing 3 ppl out of the 8 to act as unit? Like {DEF}, {AGF} etc etc.. I was thinking we would have to do 8C3 x 3! x 6!... can you tell me where I am going wrong with my logic? Thanks for your help!
I'll try to give you my interpretation, waiting for some math expert Choosing 3 people out of 8, you would calculate all the possible sub-group of 3 people you could select from a group of 8. E.g., ABC, ABD, ABE, BCD, FGH, FBE, ... so on so forth.
The question specify "if 3 of them must sit together". Therefore it is not asking to find all the possible combinations in which 3 people can always sit next to each other. Paraphrasing, it is just asking "no matter who, consider that there are 3 people that decided they must always sit together (either ABC or ABD or ABE etc..): in this case, how many combinations can we create?" . That is why you have to do your calculation only on one possible sub-group, not on all of them.
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65% (01:03) correct 
