Oneyde and Jorge must make two reservations for an airplane trip, in w

Question

GMATH  practice exercise (Quant Class 18)

https://i.postimg.cc/W31dkS9T/01-Mar19-12b.gif

Oneyde and Jorge must make two reservations for a trip, in which only 9 seats are still available, all of them in row F (see figure). If they want to sit next to each other, without being separated by central corridors, in how many ways this could be done if Oneyde refuses to sit in any of the two places next to Lilly´s seat?

(A) 9
(B) 11
(C) 13
(D) 15
(E) 17

Chethan92 · Answer

Oneyde and Jorge can sit only 1 way in which lilly is sitting.
In the first compartment, both can sit in 6 ways.
In the middle compartment, both can sit in 2 ways.
Total = 6+2+1 = 9

A is the answer.

fskilnik · Answer

?\,\,\,\,:\,\,\,\# \,\,\,O,J\,\,{\rm{positions}}\,\,{\rm{with}}\,\,{\rm{restrictions}}

{\rm{4 - seats}}\,\,{\rm{group}}\,\,{\rm{(left)}}\,\,{\rm{:}}\,\,{\rm{6}}\,\,{\rm{possibilities}}\,\,\,\left\{ \matrix{
 \,\left( {O,J,e,e} \right)\,\,{\rm{or}}\,\,\left( {J,O,e,e} \right) \hfill \cr 
 \,\left( {e,O,J,e} \right)\,\,{\rm{or}}\,\,\left( {e,J,O,e} \right) \hfill \cr 
 \left( {e,e,O,J} \right)\,\,{\rm{or}}\,\,\left( {e,e,J,O} \right) \hfill \cr} \right.\,\,\,\,\,\,\,\,\,\left

{\rm{2 - seats}}\,\,{\rm{group}}\,\,{\rm{:}}\,\,{\rm{2}}\,\,{\rm{possibilities}}\,\,\,\left\{ \matrix{
 \,\left( {O,J} \right) \hfill \cr 
 \,\left( {J,O} \right) \hfill \cr} \right.

{\rm{4 - seats}}\,\,{\rm{group}}\,\,{\rm{(right)}}\,\,{\rm{:}}\,\,{\rm{1}}\,\,{\rm{possibility}}\,\left( {O,J,L,e} \right)\,\,\,\,\,\,\left \,\,\,

? = 6 + 2 + 1 = 9

The correct answer is (A).

We follow the notations and rationale taught in the  GMATH  method.

Regards, 
Fabio.

Doer01 · Answer

Hi, 
  Considering Onedye & Jorge as one. Similarly consider two seats as 1. 
possibilities in first column = choose 1 from 3 * 2(Onedye & Jorge can switch seats, since they are considered as 1) = 3C2 * 2 = 6
Similarly in second column = choose 1 from 1 * 2 = 1C1 * 2 = 2 
When it comes to third column, there is only one way to sit = 1
total combinations = 6 + 2 + 1 = 9

Option A

Thanks.

sayan640 · Answer

In the first compartment , both can sit in 3! = 6 ways ( As  O and J need to be taken as one entity hence total 3 entities are there; and they can be arranged in 6 ways). In the second compartment , they can sit 2! = 2 ways . And In the third compartment , only one way.
Total = 6 + 2+ 1 = 9 ways.. MartyMurray

MartyMurray · Answer

Your method for the first section doesn't work since the two empty seats are the same. So, with O and J considered 1 element, you have 3 elements, but they can be arranged in only 3 ways.

OJ _ _

_ OJ _

_ _ OJ

In other words, 3!/2!.

So, you need to figure out why there are still 6 arrangements for the first section.

bumpbot · Answer

Automated notice from GMAT Club BumpBot: 

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

 This post was generated automatically.

Oneyde and Jorge must make two reservations for an airplane trip, in w

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

Oneyde and Jorge must make two reservations for an airplane trip, in w

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards