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Eight friends go to watch a movie but only 5 tickets were available.

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Eight friends go to watch a movie but only 5 tickets were available. [#permalink] New post   24 Jun 2015, 10:47
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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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B
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Re: Eight friends go to watch a movie but only 5 tickets were available. [#permalink] New post   24 Jun 2015, 11:16
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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. [#permalink] New post   24 Jun 2015, 11:55
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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 ways

The 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 seat

The 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 seat

The 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 seat

The 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 seat

i.e. Total Ways to fill five Vacant seats = 8 x 7 x 6 x 5 x 4

Answer: Option
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B
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Re: Eight friends go to watch a movie but only 5 tickets were available. [#permalink] New post   24 Jun 2015, 10:56
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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.
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Re: Eight friends go to watch a movie but only 5 tickets were available. [#permalink] New post   25 Jun 2015, 06:08
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Or we can use the slot method, trying to arrange the 8 tickest among 5 people:

A-B-C-D-E-F-G-H - 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. [#permalink] New post   21 Sep 2018, 09:02
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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. [#permalink] New post   21 Sep 2018, 09:53
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. [#permalink] New post   21 Sep 2018, 12:33
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Hey dave13

What Bunuel has merely done is given you two ways of solving the same problem.

As you may already know, nCr = {n!}/{(n-r)!*r!}

Similarly, nPr = {n!}/{(n-r)!}

So, nCr * r! = nPr and that is exactly what he has done.

Hope that helps you.
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Re: Eight friends go to watch a movie but only 5 tickets were available. [#permalink] New post   08 Mar 2019, 08:05
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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


This is a permutation problem because the seating order is important. The number of ways for 8 friends to sit and watch a movie to which only 5 of them can obtain an admission ticket is 8P5 = 8 x 7 x 6 x 5 x 4.

Answer: B
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Eight friends go to watch a movie but only 5 tickets were available. [#permalink] New post   10 Mar 2023, 11:57
Hello, I was wondering if someone could explain why the order is important in this question.

Say 8 friends be A,B,C,D,E,F,G,H and out of those only 5 can watch movie.. so a combination of ABCDE , BACDE, CDEAB etc.. are all as same 5 ppl considering as 1 combination.

Hence, I chose 8C5 cause order not matter. We are selecting 5 folks out of 8.

Can someone please explain?
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Eight friends go to watch a movie but only 5 tickets were available. [#permalink]
10 Mar 2023, 11:57
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