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Eight friends go to watch a movie but only 5 tickets were available.
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24 Jun 2015, 09:47
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57% (01:03) correct 43% (00:42) wrong based on 236 sessions
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Eight friends go to watch a movie but only 5 tickets were available. In how many different ways can 5 people sit and watch the movie? A) 8C5 B) 8*7*6*5*4 C) 5! D) \(\frac{8!}{5!}\) E) 8*5
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Re: Eight friends go to watch a movie but only 5 tickets were available.
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24 Jun 2015, 09:56



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Re: Eight friends go to watch a movie but only 5 tickets were available.
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24 Jun 2015, 10:16
Step1 : Number of ways to select 5 out of 8 is 5C8. Step2: Now above 5 can be seated in 5! ways. Therefore total = 8C5 * 5! Answer is B !!



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Eight friends go to watch a movie but only 5 tickets were available.
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24 Jun 2015, 10:55
Patronus wrote: Eight friends go to watch a movie but only 5 tickets were available. In how many different ways can 5 people sit and watch the movie? A) 8C5 B) 8*7*6*5*4 C) 5! D) \(\frac{8!}{5!}\) E) 8*5 Just Explaining in the way that is called "Fundamental Principle of Counting"There are 5 vacant seats and 8 people available to sit on them. The first place to be occupied has 8 ways in which It can be filled in Different waysThe Second place to be occupied has 7 ways in which It can be filled in Different ways because one of the 8 friends has already occupied one seatThe Third place to be occupied has 6 ways now in which It can be filled in Different ways because Two of the 8 friends have already occupied two seatThe Forth place to be occupied has 5 ways now in which It can be filled in Different ways because Three of the 8 friends have already occupied two seatThe Fifth place to be occupied has 4 ways now in which It can be filled in Different ways because Four of the 8 friends have already occupied two seati.e. Total Ways to fill five Vacant seats = 8 x 7 x 6 x 5 x 4Answer: Option
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Re: Eight friends go to watch a movie but only 5 tickets were available.
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25 Jun 2015, 05:08
Or we can use the slot method, trying to arrange the 8 tickest among 5 people:
ABCDEFGH  these are the 8 tickets
__..__..__..__..__  these are the 5 available slots for the 8 tickets.
For the first slot we have 8 options and we delete A. For the second slot we have 7 options and we delete B. For the third slot we have 6 options and we delete C. For the fourth slot we have 5 options and we delete D. For the fifth slot we have 4 options and we delete E.
All together: 8..7..6..5..4 > 8*7*6*5*4 ANS B



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Re: Eight friends go to watch a movie but only 5 tickets were available.
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21 Sep 2018, 08:02
Patronus wrote: Eight friends go to watch a movie but only 5 tickets were available. In how many different ways can 5 people sit and watch the movie? A) 8C5 B) 8*7*6*5*4 C) 5! D) \(\frac{8!}{5!}\) E) 8*5 It is clear what answer the writer wants on this, but the actual question is ambiguous. Any 5 people can sit and watch the movie in 5! ways. If you want to know how any 5 of the 8 people can sit, that is the answer we are giving. The question should be more specific.



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Eight friends go to watch a movie but only 5 tickets were available.
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21 Sep 2018, 08:53
Bunuel wrote: Patronus wrote: Eight friends go to watch a movie but only 5 tickets were available. In how many different ways can 5 people sit and watch the movie? A) 8C5 B) 8*7*6*5*4 C) 5! D) \(\frac{8!}{5!}\) E) 8*5 Choosing 5 out of 8, when order matters \(P^5_8\), or choosing 5 out of 8, without order and then arranging: \(C^5_8*5!\). Answer: B. Bunuel why didn't you use only one method , when order doesn't matter ? pls explain pushpitkc please help bunuel went to sleep



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Eight friends go to watch a movie but only 5 tickets were available.
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21 Sep 2018, 11:33
Hey dave13What Bunuel has merely done is given you two ways of solving the same problem. As you may already know, nCr = {n!}/{(nr)!*r!} Similarly, nPr = {n!}/{(nr)!} So, nCr * r! = nPr and that is exactly what he has done. Hope that helps you.
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Eight friends go to watch a movie but only 5 tickets were available. &nbs
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