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21 Nov 2024, 01:30
Kudos
Bookmarks
C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
68% (01:27) correct
32%
(01:37)
wrong
based on 230
sessions
History
Date
Time
Result
Not Attempted Yet
If 8 students are to be seated in a row of 8 seats, what is the probability that two particular students are seated at the two extreme ends of the row?
A. 1/112
B. 1/56
C. 1/28
D. 1/14
E. 1/7
A. 1/112
B. 1/56
C. 1/28
D. 1/14
E. 1/7
21 Nov 2024, 07:34
Kudos
Bookmarks
Our formula: Favorable Arrangements
-----------------Possible Arrangements
Total Possible Arrangements:
8 students in 8 seats:
8! = 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1
Possible Favorable Arrangements:
-two students are seated at both ends. Let's call them student A & B.
We have 2 possible favorable outcomes.
Along with the two favorable outcomes, we have 6 students who can be placed in the remaining 6 seats. This is represented as 6!.
2(end arrangements) x 6!(the students sat in-between) = (2) x (6!)
Going back to our original formula:
-2 x 6! = --2 x6 x 5 x 4 x 3 x 2 x 1--
---8!------8 x 7 x6 x 5 x 4 x 3 x 2 x 1
= 2/56 = 1/28
C
-----------------Possible Arrangements
Total Possible Arrangements:
8 students in 8 seats:
8! = 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1
Possible Favorable Arrangements:
-two students are seated at both ends. Let's call them student A & B.
- Student A is seated at far left. Student B is seated at far right.
- Student B is seated at far left. Student A is seated at far right.
We have 2 possible favorable outcomes.
Along with the two favorable outcomes, we have 6 students who can be placed in the remaining 6 seats. This is represented as 6!.
2(end arrangements) x 6!(the students sat in-between) = (2) x (6!)
Going back to our original formula:
-2 x 6! = --2 x
---8!------8 x 7 x
= 2/56 = 1/28
C
21 Nov 2024, 13:21
Kudos
Bookmarks
Bunuel
The total outcome will be 8P8 = 8!
To calculate the favourable outcomes,
- - - - - - - -
Among them, specific 2 will be seated on extreme ends in any 2! ways
The remaining seats are 6, so the remaining 6 people will be seated in 6! ways
Hence probability will be (2!6!)/8!= 2/56= 1/28
