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

Question

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

GMATinsight · Accepted Answer

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  B

NickHalden · Answer

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 !!

Bunuel · Answer

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.

pacifist85 · Answer

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

Music2mba · Answer

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.

dave13 · Answer

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

pls explain

pushpitkc please help

bunuel went to sleep

pushpitkc · Answer

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.

ScottTargetTestPrep · Answer

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

BusinessAdmin · Answer

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?

Nikkz99 · Answer

If Question asked No: of ways then 8C5 it specifically asked for No: of different ways so, order matters! therefore permutation is way to go

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

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