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

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Current Student
Joined: 21 Aug 2014
Posts: 138
GMAT 1: 610 Q49 V25
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Eight friends go to watch a movie but only 5 tickets were available.  [#permalink]

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24 Jun 2015, 10:47
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Difficulty:

45% (medium)

Question Stats:

59% (01:03) correct 41% (00:45) wrong based on 219 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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Joined: 02 Sep 2009
Posts: 50001
Re: Eight friends go to watch a movie but only 5 tickets were available.  [#permalink]

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24 Jun 2015, 10:56
1
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: $$C^5_8*5!$$.

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Re: Eight friends go to watch a movie but only 5 tickets were available.  [#permalink]

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24 Jun 2015, 11:16
2
1
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!
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Eight friends go to watch a movie but only 5 tickets were available.  [#permalink]

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24 Jun 2015, 11:55
2
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

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Re: Eight friends go to watch a movie but only 5 tickets were available.  [#permalink]

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25 Jun 2015, 06:08
1
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]

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21 Sep 2018, 09:02
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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Joined: 09 Mar 2016
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Eight friends go to watch a movie but only 5 tickets were available.  [#permalink]

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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: $$C^5_8*5!$$.

Bunuel why didn't you use only one method , when order doesn't matter ? pls explain

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Eight friends go to watch a movie but only 5 tickets were available.  [#permalink]

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21 Sep 2018, 12:33
1
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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Eight friends go to watch a movie but only 5 tickets were available. &nbs [#permalink] 21 Sep 2018, 12:33
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