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a) Three families celebrate Thanksgiving together: Alvarados, the Bells, and the Carsons. There are four people in each family. Afterwards they line up for a large group photo. How many ways can the 12 of them line up?

b) this time they want to do it so that the Alvarados come first (as you scan from left to right), then the Bells, then the Carsons.Now how many ways can they line up?

c) This time they are going to keep the families together but it doesn't matter what order the families appear (so this time you could have all of the Carsons, followed by all of the Alvarados and then all of the Bells). Now how many ways can they do it?

off-topic
 Nov 18, 2019
 #3
avatar+105476 
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a)  12!  ways  =  479, 001, 600 ways

 

b )  There are  4! ways to arrange each family = 24 so  24 * 24 * 24  = 24^3  =   13824 ways

 

c)  Assuming that each fanily can be arranged in any order....we have  6 ways to order the families  * the  ways to arrange the people within each family so   6 * 13824  = 82944 ways

 

Edit to correct my last two answers....

 

 

cool cool cool

 Nov 18, 2019
edited by CPhill  Nov 18, 2019
 #4
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Thank for your help. Can you explain the question (a) How  12!  ways  =  479, 001, 600 ways? Thank you.

Guest Nov 18, 2019
 #5
avatar+105476 
+1

We have 12 ways to choose the first person in line

 

11 ways to choose the second

 

10 to choose the third....etc...

 

So

 

12 * 11 * 10 * 9 *.........* 3 * 2 * 1      =   12!

 

 

cool cool cool

CPhill  Nov 18, 2019
 #6
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Im sorry cphil but 2nd answer is incorrect

Guest Nov 18, 2019
 #7
avatar+105476 
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Corrected...

 

 

cool cool cool

CPhill  Nov 18, 2019
 #8
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Thank you,

 Nov 18, 2019
edited by Guest  Nov 18, 2019

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