In how many ways can you seat 7 people on a bench if one of

Question

In how many ways can you seat 7 people on a bench if one of them, Rohit, does not want to sit on the middle seat or at either end?

A) 720
B) 1720
C) 2880
D) 5040
E) 10080

Bunuel · Accepted Answer

Since Rohit does not want to sit on the middle seat or at either end (3 chairs), then he can choose 4 chairs to sit. The remaining 6 people can sit in 6! ways. Thus the # of arrangements is 4*6! = 2,880.

Answer: C.

knicks1288 · Answer

There are 7! total ways to sit everyone.

There are 3 seats that he can't sit in (the two ends and the middle). For each seat of the 3 seats he can't sit in, there are 6! arrangements of people in the other seats. So the total number of arrangements where he isn't sitting in any of those three seats is:

7! - 6!(3) = 
6!(7 - 3) = 
6!(4) = 
2880

nick28 · Answer

total no of ways in which all 7 can be seated =  7! ways

now, 
 if rohit sits at  first  seat then remaining 6 can be seated in  6! ways 
 if rohit sits at  middle  seat then remaining 6 can be seated in  6! ways 
 if rohit sits at  last  seat then remaining 6 can be seated in  6! ways

no. of ways in which rohit  doesnt  sits in middle or at either ends =  7! - (6! + 6! + 6!)  = 7! - 6!*3 = 6!(7-3)=6!*4= 2880

Hope this helps.

Archit3110 · Answer

total seating arragements = 7!
for Rohit who supposedly sits on first chair ; then 6! ways for others
middle seat : again 6! and last seat 6!

so total ways in which all of them can sit 
7!-3*6!
6! ( 7-3)
6!*4
= > 2880 
IMO C

ScottTargetTestPrep · Answer

Solution:
 
If there is no restriction on the seating arrangement, there will be 7! = 5040 ways to seat the 7 people.

If Rohit must sit on the middle seat, there will be 6! = 720 arrangements. However, since he can’t sit there, we must subtract 720 from 5040. Likewise, since he can’t sit at either end, we have to further subtract 720 twice (once for each end) from 5040. In other words, we have to subtract 3 times 720 from 5040. Therefore, there are 5040 - 3(720) = 5040 - 2160 = 2880 seating arrangements if Rohit does not want to sit on the middle seat or at either end.

Alternate Solution:

For the first seat, there are 6 options since Rohit will not sit on the first seat. For the fourth seat (which is the middle seat), there are 5 options since we must exclude Rohit and the person chosen for the first seat. Likewise, there are 4 options for the last seat since we are excluding Rohit and the two previously selected people.

For the second seat, there are also 4 options since we now include Rohit. For the third, fifth and sixth seats, there are 3, 2, and 1 available options, respectively. Thus, the number of arrangements for the 7 people to sit as described in the question is 6 x 5 x 4 x 4 x 3 x 2 x 1 = 2,880.

Answer: C

GMATinsight · Answer

7 seats for 7 people are _ _ _ _ _ _ _

Where Rohit can't sit are x _ _ x _ _ x

i.e. Options of Rohit's seat = 4 (remaining dashes)

Remaining 6 people can sit on 6 places in 6! ways = 720 ways

Total seating arrangements = 4*6! = 2880

Answer: Option C

CEdward · Answer

Permutations with restrictions:

7! < ---- # of ways without restrictions
6! <--- # of ways where Rohit is fixed in the center
6! <--- # of ways where Rohit is at one end only
6! <---- # of ways where Rohit is at the other end

7! - (6! x 3 = 720 x 3 = 2160) = 7! - 2160 + X

X = 5040 - 2160 = 2880

Answer is C.

chiplesschap · Answer

Tried a similar approach using basic counting principle but couldn't get to the right answer. I did as follows :

Assumption -> A, B, C, D, E, F & Rohit as the 7 person

Seat-1 : A, B, C, D, E & F -> 6
Seat 2 : 7 (as Rohit can occupy Seat 2) 'minus' occupant of Seat-1 -> 6
Seat 3 : 7 (as Rohit can occupy Seat 3) 'minus' occupants of Seat-1 & 2 -> 5
Seat 4 : 6 (as Rohit cannot occupy Seat 4) 'minus' occupants of Seat-1, 2 & 3 -> 3
Seat 5 : 7 (as Rohit can occupy Seat 5) 'minus' occupants of Seat-1 to 4 -> 3
Seat 6 : 7 (as Rohit can occupy Seat 5) 'minus' occupants of Seat-1 to 5 -> 2
Seat 7 : 6 (as Rohit cannot occupy Seat 7) 'minus' occupants of Seat-1 to 6 -> 1

POSSIBLE MISTAKE :

I think the right approach for such questions is seats with constraints i.e. Seats - 1, 4 & 7 should first be filled. It then becomes a straightforward counting principle question.

Although the alternate approach i.e. "Total possible outcomes" - "Unfavourable outcomes" is probably the fastest way to solve.

Total possible outcomes of seating 7 persons -> 7!

Unfavourable outcomes -> Rohit occupying Seat-1 -> 6! + Rohit occupying Seat-2 -> 6! + Rohit occupying Seat-3 -> 6!

=> 7!-3x6! = 2880

Thanks!

pal26 · Answer

How I solved this.

Select 3 seats (Middle seat and two end seats) for 6 people. (6 because excluding Rohit) . 6C3.

These 3 people seating can be arranged in 3! ways. So, 6C3*3!.

Select rest 4 seats for remaining 4 people and arranging them. 4C4 * 4!.

Total no of ways= 6C3*3!*4C4*4!= 2880. Hence C.

In how many ways can you seat 7 people on a bench if one of

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

In how many ways can you seat 7 people on a bench if one of

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