Six friends go to watch a movie. They are supposed to occupy seat numbers 51 to 56. However one of them falls sick and r

BTGModeratorVI wrote: ↑

Wed Jan 06, 2021 7:59 am

Six friends go to watch a movie. They are supposed to occupy seat numbers 51 to 56. However one of them falls sick and returns home. In how many different ways can the 5 people sit?

A. 240
B. 480
C. 600
D. 720
E. 960

Answer: D

Solution:

Let the 5 people who are present at the movie theater be A, B, C, D and E and let _ denote the empty seat. So we could have:

_ABCDE, A_BCDE, AB_CDE, ABC_DE, ABCD_E and ABCDE_

However, for each of these 6 seating arrangements, there are 5! ways to arrange the 5 people while keeping the empty seat where it is. For example, for A_BCDE (i.e., the second seat is empty), we could have C_DABE, D_EBAC, etc.

Therefore, there are a total of 6 x 5! = 6 x 120 = 720 seating arrangements.

Alternate Solution:

We can let the 5 people be represented by A, B, C, D, and E, and the empty seat/sixth person be designated as F. Note that there is no difference between an “empty” seat and one occupied by the original ticket purchaser. Thus, we have 6 “people” to be seated in 6 seats. This results in there being 6! = 720 different seating arrangements.

Answer: D

BTGModeratorVI wrote: ↑

Wed Jan 06, 2021 7:59 am

Six friends go to watch a movie. They are supposed to occupy seat numbers 51 to 56. However one of them falls sick and returns home. In how many different ways can the 5 people sit?

A. 240
B. 480
C. 600
D. 720
E. 960

Answer: D
Source: Veritas Prep

We must assign each of the 5 remaining friends to one of the 6 seats.
So, the 1st friend has 6 options
Next, the 2nd friend has 5 options remaining
The 3rd friend has 4 options
The 4th friend has 3 options
And 5th friend has 2 options
TOTAL number of ways to sit the 5 people = (6)(5)(4)(3)(2) = 720

Answer: D