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Difficulty:
35%
(medium)
Question Stats:
69% (01:11) correct
31%
(01:18)
wrong
based on 900
sessions
History
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Not Attempted Yet
A music teacher is preparing for a recital in which 3 pairs of students, pair P, pair Q, and pair R, will perform duets. The teacher must decide in what order the 3 pairs will perform and in what order the 2 students in each pair will be introduced immediately before they perform. How many different orderings of student introductions resulting from these decisions are possible?
A. 6
B. 12
C. 48
D. 120
E. 720
A. 6
B. 12
C. 48
D. 120
E. 720
17 Aug 2023, 12:26
Kudos
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12922
A music teacher is preparing for a recital in which 3 pairs of students, pair P, pair Q, and pair R, will perform duets. The teacher must decide in what order the 3 pairs will perform and in what order the 2 students in each pair will be introduced immediately before they perform. How many different orderings of student introductions resulting from these decisions are possible?
A. 6
B. 12
C. 48
D. 120
E. 720
A. 6
B. 12
C. 48
D. 120
E. 720
The teacher has 4 decisions to make.
The number of ways the 3 pairs can be arranged is 3!.
The number of ways the 2 students in pair P can be arranged is 2!.
The number of ways the 2 students in pair Q can be arranged is 2!.
The number of ways the 2 students in pair R can be arranged is 2!.
So, the number of ways the teacher can make all 4 decisions is:
3! × 2! × 2! × 2! = 6 × 2^3 = 6 × 8 = 48
Answer: C
15 Jun 2023, 01:17
Kudos
Bookmarks
12922
A music teacher is preparing for a recital in which 3 pairs of students, pair P, pair Q, and pair R, will perform duets. The teacher must decide in what order the 3 pairs will perform and in what order the 2 students in each pair will be introduced immediately before they perform. How many different orderings of student introductions resulting from these decisions are possible?
A. 6
B. 12
C. 48
D. 120
E. 720
A. 6
B. 12
C. 48
D. 120
E. 720
The 3 pairs can perform in 3! = 6 ways.
The 2 students in each pair can be introduced in 2 ways.
Therefore, the total number of different orderings of student introductions is 6*2*2*2 = 48.
Answer: C.
