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In how many ways can you sit 8 people on a bench if 3 of

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In how many ways can you sit 8 people on a bench if 3 of [#permalink]

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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
[Reveal] Spoiler: OA

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Re: Combinations [#permalink]

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New post 12 Mar 2012, 09:35
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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


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.
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Re: In how many ways can you sit 8 people on a bench if 3 of [#permalink]

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New post 12 Mar 2012, 09:59
thanks!

did you consider following arrangements?
1: -{D},{ABC}, {E}, {F}, {G}, {H},
2: -{D},{E}, {F},{ABC}, {G}, {H},
3: -
..

Please clarify.

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Re: In how many ways can you sit 8 people on a bench if 3 of [#permalink]

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New post 12 Mar 2012, 10:30
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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.

Check Combinations chapter of Math Book for more: math-combinatorics-87345.html

Hope it helps.
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Re: In how many ways can you sit 8 people on a bench if 3 of [#permalink]

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New post 29 May 2016, 23:57
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.

Total ways: 6!*3! = 4320.

D is the answer.

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Re: In how many ways can you sit 8 people on a bench if 3 of [#permalink]

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New post 24 Oct 2017, 08:35
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!

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In how many ways can you sit 8 people on a bench if 3 of [#permalink]

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New post 29 Oct 2017, 12:20
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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.

Hope it is clear...and correct! :)

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Re: In how many ways can you sit 8 people on a bench if 3 of [#permalink]

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New post 31 Oct 2017, 16:40
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


Since we are not given any names, we can denote each person with a letter:

A, B, C, D, E, F, G, H

Let’s say A, B, and C must sit together; we treat [A-B-C] as a single entity, and so we have:

[A - B - C] - D - E - F - G - H

We see that we have 6 total positions, which can be arranged in 6! = 720 ways. We also can organize [A - B - C] in 3! = 6 ways.

So, the total number of ways to arrange the group is 720 x 6 = 4,320 ways.

Answer: D
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Re: In how many ways can you sit 8 people on a bench if 3 of   [#permalink] 31 Oct 2017, 16:40
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