Problem Solving

Six People - Alex, Bryan, Chan, Dan, Emily and Fabian - were the only invitees to an award ceremony. On the day of the ceremony, it was informed that Chan will not be able to attend the meeting. In how many ways can the remaining people be seated on the six chairs such that Bryan sits on the left of Fabian?

A) 48
B) 120
C) 288
D) 360
E) 720

OA - D

Here's what I did:

Method 1 - Out of the six possible chairs, there could be five chairs for Brian ( we have to leave the right most chair for Fabian)

Brian - 5 possibilities
Fabian - 5 possibilities (he cannot occupy the left most chair)

Dan, Alex, Emily can occupy "any" of the 4,3,2 seats respectively.

Hence, the total number of combinations = 5 * 5 * 4 * 3 * 2 = 600. Unfortunately, that's incorrect.

Method 2 -

Let's assume that B and F sit together = BF. Hence, we have four people, BF, A, D and E. They can be arranged in 4P6 ways = 360. I know that the answer is correct, but my method is incorrect because B and F don't have to necessarily sit right next to each other.

Any thoughts?

Thanks

First of all its a very nice question and you should refer Dr.R.D.Sharma's book for the best answer of this question because It contains very nice explanation and examples also.
Cylinder Volume

Ok, let me try to point out the mistake.

Method 1:
You're right in saying that there are 5 possibilities for Bryan.
So, the first '5' is correct.

Say Bryan occupied the 3rd seat from the left. If this happens, Fabian DOES NOT have 5 possibilities. He just has has three possibilities.
If Bryan occupies the 5th seat from the left, Fabian has just one possibility.

Do you get the drift now?

If Bryan has 5 options, Fabian does not necessarily have 5 options. The number of options that Fabian varies depending on where Bryan sits.

So, the second '5' is wrong.

Let's solve by this method in a correct manner and see if we can the correct answer.

Bryan sits on the 1st Seat => Fabian has 5 options.
Bryan sits on the 2st Seat => Fabian has 4 options.
Bryan sits on the 3st Seat => Fabian has 3 options.
Bryan sits on the 4st Seat => Fabian has 2 options.
Bryan sits on the 5st Seat => Fabian has 1 options.

Total number of ways to make Bryan and Fabian sit like this
= 1*5 + 1*4 + 1*3 + 1*2 + 1*1
= 15

Required number of ways
= 15*4*3*2
= 360

Method 2:
For arranging (BF), A, D and E, 6P4 is incorrect because (BF) occupy two adjacent seats. (BF), A, D and E have 5 seats to sit on. There are a few more mistakes as well, but let's stop the discussion here.
Basically, you were just plain lucky to get the right answer.

The others can view the Video Explanation of the better solution here:
https://www.youtube.com/watch?v=ZXL3KnRu51A&feature=plcp

First of all this is a counting problem and not probability
(just nit-picking )

Here is my solution. I'm not sure of the time consumption part
though. Maybe others have a better solution.

Absence of Chan does not matter in this problem. Hence let us put
Chan back in the problem. Without any constraint, the total number
of arrangements would have been 6 * 5 * 4 * 3 * 2 * 1. But, with the
constraint, below are the possible arrangements.

If the positions are numbered

1 2 3 4 5 6,

then

Brian in position 1 => 5 * 4 * 3 * 2 * 1 = 120

Brian in position 2 => 3, 4, 5, 6 can be arranged in 24 (4 * 3 * 2 * 1)
ways. But position 1 can change in 4 ways.

Hence total (4 * 24) = 96

Brian in position 3 => 4, 5, 6 can be arranged in 6 ways (3 * 2 * 1), But
positions 1 and 2 can be arranged in (4 * 3).

Hence total (12 * 6) = 72

Brian in position 4 => 5,6 in 2 ways. 1, 2, 3 in (4 * 3 * 2).

Hence total (24 * 2) = 48

Brian in position 5 => 6 in 1 way. 1, 2, 3, 4 in (4 * 3 * 2 * 1).

Hence total (24 * 1) = 24

Total = 360

HTH

voodoo_child wrote:Six People - Alex, Bryan, Chan, Dan, Emily and Fabian - were the only invitees to an award ceremony. On the day of the ceremony, it was informed that Chan will not be able to attend the meeting. In how many ways can the remaining people be seated on the six chairs such that Bryan sits on the left of Fabian?

A) 48
B) 120
C) 288
D) 360
E) 720

OA - D

Here's what I did:

Method 1 - Out of the six possible chairs, there could be five chairs for Brian ( we have to leave the right most chair for Fabian)

Brian - 5 possibilities
Fabian - 5 possibilities (he cannot occupy the left most chair)

Dan, Alex, Emily can occupy "any" of the 4,3,2 seats respectively.

Hence, the total number of combinations = 5 * 5 * 4 * 3 * 2 = 600. Unfortunately, that's incorrect.

Method 2 -

Let's assume that B and F sit together = BF. Hence, we have four people, BF, A, D and E. They can be arranged in 4P6 ways = 360. I know that the answer is correct, but my method is incorrect because B and F don't have to necessarily sit right next to each other.

Any thoughts?

Thanks

amit28it wrote:First of all its a very nice question and you should refer Dr.R.D.Sharma's book for the best answer of this question because It contains very nice explanation and examples also.
Cylinder Volume

Thanks Amit...I don't have my books with me right now. They are in India. It's been quite a few years since I did these problems in Class 12th/pre-NIT days. I know that I was weak in Probability during those days. Probability questions that we used to get those days were more about Bernoulli's, conditional prob, Baye's theorem, Correlation coefficient etc. These are easy. However, now my brain doesn't cooperate on these problems.... haha

Hi Mike,

This is a problem designed by me, so when I was making the problem I deliberately wrote 'Bryan sits on the left of Fabian' so that people are misled into believing that Bryan sits immediately to the left of Fabian. Option (B): 120 was placed for those who fall into this trap.

I am not an expert at designing problems, but I still think that if the problem says that 'Bryan sits on the left of Fabian', it means that Bryan can be anywhere on the left of Fabian.

Regards
Aneesh

P.S.:
The video explanation is here:
https://www.youtube.com/watch?v=ZXL3KnRu51A

Let me know if you have any feedback.

The answer 360 is incorrect, it is 60!
120 is also incorrect if you assume Bryan sits immediately left of Fabian, it is 24.

Also, I'm sorry, but tricking someone with unclear wording is a poor trick, especially when the math should be the trick.

Let's look at this:

If this problem has NO constraints and people can sit in any chair, then it is solved as follows:

Chair 1 = > 5 people can sit here
Chair 2 = > 4 people, 1 already sitting
Chair 3 = > 3 people, 2 already sitting
Chair 4 = > 2 people, 3 already sitting
Chair 5 = > 1 person, 4 already sitting

The numbers decrease each chair since someone can obviously not sit in 2 chairs at once.
So: 5 x 4 x 3 x 2 x 1 = 5! = 120 <== MAXIMUM possible amount for a 5 set permutation with no repeating allowed.

120 is the MAXIMUM number of permutations available. Bryan can sit in chair 5 and Fabian can sit in chair 1 here, both options that are unavailable on the real problem. Rules REDUCE this number!

The problem introduces a constraint, "Bryan sits on the left of Fabian", which will REDUCE the number of available options, not raise it to 360!!!

Since the Author responded and said he really meant "Bryan sits in any chair to the left of Fabian", I will solve it that way first.

So:
Bryan in Chair 1, Fabian has 4 options
Bryan in Chair 2, Fabian has 3 options
Bryan in Chair 3, Fabian has 2 options
Bryan in Chair 4, Fabian has 1 options

This would be 10 different Bryan/Fabian configurations, so 10 x 3 x 2 x 1 = 60!

In detail:

Each of these has 1 chair constrained to Bryan. Let's look at "Bryan in Chair 1, Fabian has 4 options"

That would be:
Chair 1 = 1 for Bryan
Chair 2 = 4 people can sit here, including Fabian
Chair 3 = 3 people can sit here, 2 already sitting
Chair 4 = 2 people can sit here, 3 already sitting
Chair 5 = 1 person can sit here, 4 already sitting

So: 1 x 4 x 3 x 2 x 1 = 24 permutations with Bryan in chair 1

Now, put Bryan in chair 2. Chair 1 can only have 3 people in it with Brian or Fabian constrained out of it.
Chair 1 = 3 here because of 5 people minus Bryan and Fabian
Chair 2 = 1 for Bryan
Chair 3 = 3 here, 2 already sitting
Chair 4 = 2 here, 3 already sitting
Chair 5 = 1 person can sit here
So, 3 x 1 x 3 x 2 x 1 = 18 permutations!

Keep this going, Brian in Chair 3, Fabian restricted to chair 4,5
3 x 2 x 1 x 2 x 1 = 12 permutations

Brian Chair 4, Fabian MUST BE chair 5 so,
3 x 2 x 1 x 1 x 1 = 6 permutations

Total: 24 + 18 + 12 + 6 = 60

If the problem was "Bryan must sit immediately to the left of Fabian", then the answer is 24.
You constrain two seats to 1 so that ( 1 x 1 x 3 x 2 x 1) = 6. The 2 seats constrained can be in 4 spots:
1 1 _ _ _
_ 1 1 _ _
_ _ 1 1 _
_ _ _ 1 1

so, (6 x 4 = 24)
OR, 4 x 3 x 2 x 1 = 24

gmat_and_me wrote:First of all this is a counting problem and not probability
(just nit-picking )

You mean permutation, right?

Hi Shape,

I just want to mention a couple of things.

First, There are 6 seats and 5 people have to be arranged on them in a specified manner. Your answer gets multiplied with a 6C5 (or a 6) because we can arrange these 5 people on any 5 of the 6 available seats.

Second, Playing with words is not a poor trick at all. GMAT plays around with words all the time. You always have to read the language of the problem very carefully.
We can debate on whether the wording was clear or not but I am reasonably confident that if B sits on the left of F, he can be anywhere on the left of F. We are so used to solving problems in which people/objects are placed together that we are misled into believing that it is the same case here.

aneesh.kg wrote:Hi Shape,

I just want to mention a couple of things.

First, There are 6 seats and 5 people have to be arranged on them in a specified manner. Your answer gets multiplied with a 6C5 (or a 6) because we can arrange these 5 people on any 5 of the 6 available seats.

Second, Playing with words is not a poor trick at all. GMAT plays around with words all the time. You always have to read the language of the problem very carefully.
We can debate on whether the wording was clear or not but I am reasonably confident that if B sits on the left of F, he can be anywhere on the left of F. We are used to solving problems in which people sit together that we are mislead into believing that it is the same case.

So are you treating the empty chair as a choice then, so it is really not like Chan left at all?

Are you treating the following orders an 1 combination or multiple?
Alex, Bryan, EMPTY, Dan, Emily, Fabian
Alex, Bryan, Dan, EMPTY, Emily, Fabian
Alex, Bryan, Dan, Emily, EMPTY, Fabian
Alex, Bryan, Dan, Emily, Fabian, EMPTY
EMPTY, Alex, Bryan, Dan, Emily, Fabian

If treating as multiple, then it is like Chan never left.

All the ways listed by you are different ways because it matters which one of the six chairs is empty; and yes, it is as though Chan was present at the ceremony.

Therefore the best solution is this:
(6!)/2

because in the 6! ways of arranging six different things, in 50% of the ways B will be before F.

I think Mike is right here.
https://www.beatthegmat.com/counting-six ... 47167.html
This link should explain the reasoning.

Hi Mike,

Point noted; and thanks for pointing out how the wording can be improved. I am not 100% convinced, but I trust your opinion because you know the GMAT better than I do.

Hi Niket,

The Mobsters' problem posted by you is really interesting and has much better wording than the one made by me.

gmat_and_me wrote:
Brian in position 2 => 3, 4, 5, 6 can be arranged in 24 (4 * 3 * 2 * 1)
ways. But position 1 can change in 4 ways.

Hence total (4 * 24) = 96

GMAT_AND_ME = I didn't get why you are saying that "position 1 can change in 4 ways"..... Same goes with Brian occupying the second position. Can you please explain this?