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
