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In how many ways can three pairs of siblings from different families be seated in two rows of three chairs, if siblings may sit next to each other in the same row, but no child may sit directly in front of their sibling?

 Mar 30, 2023
 #1
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This question has been asked a lot ot times recently.

there are probably a number of different answers given.

 

I let the children be     A1    A2    B1    B2    C1    C2

If there were no restrictions there would be 6! ways to seat them

 

How many ways can they NOT be seated.

 

We cannot have all the pairs under each other so  that would be

6 places to put A1    then 1 place to put A2

4 places to put B1   then 1 place to put  B2

2 places to put  C1  then 1 place to put C2

6*4*2 = 48

 

We cannot have one pair vertically alligned with the others not so

Say the As are vertially alligned but not B or C

6 places to put A1, 1 place to put A2

4 places to put B1, 2 places to put B2

2 places to put C1 and 1 place to put C2

6*1*4*2*2*1 = 96 ways

 

BUT ic could have been A or B or C that are vertically aligns so we have to multiply by 3

96*3=288

 

So I think the answer is   

6! - (48+288) = 6! - 336 = 720 - 336  =  384  ways

 Mar 31, 2023
 #2
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+1

Take anyone of the 6 kids:

 

He/she has 6 choices. The sibling has 5 - 1 = 4 choices.

 

The 4th kid has the remaining 4 choices. The sibling has 3 - 1=2 choices

 

The last pair has the remaining 2 choices.

 

Total seating =6  x  4  x  4  x  2  x  2 =384 ways of seating the 3 pairs of siblings.

 Mar 31, 2023

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