A teacher must seat six students in 2 rows of three seats each. If the

Question

A teacher must seat six students in 2 rows of three seats each. If the one troublesome student must sit in the front row, and the others may sit in either row, how many possible seating arrangements does the teacher have to choose from?

A. 120
B. 240
C. 300
D. 360
E. 720

A. 120
B. 240
C. 300
D. 360
E. 720

BrentGMATPrepNow · Accepted Answer

Let of the 6 students be, A, B, C, D, E and T, where T represents the troublesome student.

Take the task of seating the 6 students and break it into   stages  .

We’ll begin with the  most restrictive stage .

Stage 1 : Select a seat for child T to sit on 
We can choose any of the 3 front seats, so we can complete stage 1 in  3  ways

Stage 2 : Select a seat for child A to sit on 
There are 5 empty seats remaining, so we can complete stage 2 in  5  ways

Stage 3 : Select a seat for child B to sit on 
There are 4 empty seats remaining, so we can complete stage 3 in  4  ways

Stage 4 : Select a seat for child C to sit on 
There are 3 empty seats remaining, so we can complete stage 4 in  3  ways

Stage 5 : Select a seat for child D to sit on 
There are 2 empty seats remaining, so we can complete stage 5 in  2  ways

Stage 6 : Select a seat for child E to sit on 
There is 1 empty seat remaining, so we can complete stage 6 in  1  way

By the Fundamental Counting Principle (FCP), we can complete all 6 stages (and thus seat all 6 children) in  (3)(5)(4)(3)(2)(1)  ways (= 360 ways)

Answer: D

Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. So, be sure to learn it.

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Regor60 · Answer

There are 2 additional students to be added to the front row, given that the troublesome student is assigned there.

5!/2!3! = 10 ways to select the 2 students from the remaining 5.

There are now 5-2 = 3 students remaining for the back row, so no additional selection is required.

The 3 students in each row can be arranged 3! =6 ways

Total arrangements: 10*6*6=360

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JerryAtDreamScore · Answer

Breaking Down the Info:

We would like to start with the student with fewer options. The troublesome student must select one of the front seats, so we place that student first.

That student would have 3 options. Then we can randomly assign the rest of the 5 students.

There are 5 spaces left with 5 students, so there are  5P5 = 5!  random assignments.

Thus in total, there are  3*5! = 3*120 = 360  options.

Answer: D

Fdambro294 · Answer

Quickest way to visualize the answer:

If there were no constraints at all, we could arrange the 6 people among 6 distinct seats in:

6! = 720 ways

Call the troublesome student Scott

Because the seats are broken up evenly — 3 are in the front and 3 are in the back —

(1/2) of all the possibilities will have Scott in one of the 3 front seats

And

(1/2) of all the possibilities will have Scott in one of the 3 back seats

We only want to keep the possibilities in which Scott is in one of the 3 front seats:

(6!) * (1/2) =
(720) * (1/2) =

360

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shivani1351 · Answer

Use visualization to answer this question in the easiest possible way:
Let the 6 students be A,B,C,D,E and F, and let the troublesome student be A, who needs to be seated in the front row.
Let R1 be the front row.

Case 1: 
R1 -> A _ _
R2 -> _ _ _

In this case, the number of ways of seating them all = 5! = 120

Case 2: 
R1 -> _ A _
R2 -> _ _ _

In this case, the number of ways of seating them all = 5! = 120

Case 3: 
R1 -> _ _ A
R2 -> _ _ _

In this case, the number of ways of seating them all = 5! = 120

Hence, the total number of ways of seating them all = 120*3 = 360 (Option D)

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