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Gavan
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Gavan
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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Gavan
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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Gavan
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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Gavan
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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Gavan
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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Gavan
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

possible ways = 3! for 3 people and 6! for total group of 5 single + 1 of 3 people
6*6! = 4320
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Gavan
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


8 people if 3 have to sit together can be written as :

5+3= 6 (considering 3 as 1 batch)

6 people can sit together in 6! ways and 3 people (considered as 1 batch) can sit together in 3! ways.

total ways hence can be 6!x3!= 4320

Answer: C
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Take 3 persons as a unit. So total of 6 persons. Six people can sit in 6! Ways. 3 people can rearrange themselves in 3! Ways. So 6! 3! Ways. So 720*6= 4320 ways. Hope you understood.

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