A music teacher is preparing for a recital in which 3 pairs of student

Question

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

JeffTargetTestPrep · Accepted Answer

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

Bunuel · Answer

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.

Apk641 · Answer

This is a permutation problem where we can use P(n,r)= n!/(n−r)!

You are arranging 3 pairs of students, where order matters (which pair performs first, second, third). When order matters, we use permutations. First lets arrange the 3 pairs of students by plugging into the permutation formula.

Step 1.
P(3,3)= 3!/(3!-3!)!
=3!/0! (zero becomes 1)
=3!/1
=3*2=6

Step 2. 
Now, for each pair of students (say, A and B), there are only 2 students to arrange. The only possible arrangements are:
A goes first, B goes second
B goes first, A goes second
there are only 2 possible orders which is 2! or 2
we do this three times which = 2*2*2= 2^3=8

Multiply both results = 6*8=48

​

Dbrunik · Answer

visual:

DanTheGMATMan · Answer

Lovely ordering problem: 3! orders for P, Q, R, and 2 orders within each of the pairs, so 3! *2*2*2:
 https://www.youtube.com/watch?v=y6u81Opq4Qk

anon3612 · Answer

Initially I had used the usual method 3x2x1 to select pair order and then x8 for intro order to get 48 , but later opted for 12 based on below workings. Can somebody let me know why below approach is wrong ,
aren't we supposed to select sequentially without replacement , after choosing 1st pair , 2 pairs are left and then 1 pair.

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mbaaccount1234 · Answer

Hello, what would it mean to do 6!/(2!*2!*2!) in this example? How does it differ from doing 3!*2!*2!

Aboyhasnoname · Answer

For those wondering why not just 3! X 2...After all, the pairs can be arranged in 3! ways and then 2 possible introductions within each time...Here's the catch.... 
When you do this...you are just considering outcome/possible sequence of introduction for 1 of the pairs.......AB CD EF .....BA CD EF.......This....
But, you need the two ways of introducing for each team member....Hence 3! x 2 x 2 x2 = 48

For those wondering how to do it via nPr....Your n is 3 and your r is also 3....Now 3!/(3-3)! = 3! Gives 6...(0! = 1) 
Now arrange these pairs....nPr 2!/(2-2)! = 2..and this will happen for all three hence 2 x 2 x 2...Final answer is 3! x 2 x 2 x 2 = 48

A music teacher is preparing for a recital in which 3 pairs of student

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A music teacher is preparing for a recital in which 3 pairs of student

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