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# PTL5#23

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Jun 28, 2019

#1
+109512
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I assume you do not have the logic but do you have the answer?

Jun 28, 2019
#4
+109512
+2

You need to look at thre different cases.

1) The pairs all sit infront/behind.

See how many ways the first row can be filled and the second row will automatically be accounted for.

I think that is just 10*8*6*4*2 = 3840 possibilities.

2) 1 pair sit in each row side by side the others are infront/behind

Side by side couples:

there is 4 possible places to sit the pairs.

There are 10 ways to chose the first person of the first pair and then 8 ways to chose the first person of the second pair

So I guess that is 8*10 *4= 320 Ways.

Now there is 3 paired people/places left  6*4*2=48 ways to sit them

320*48=  15360 possibilities

3) 2 pairs sit in each row, the others are infront/behind.

The individual one can be in the 1st 3rd or 5th positions.

There are 5 ways to choose the odd pair so that make it 3*5*2= 30ways to chose.

Now there is 4 couples to sit side by side. and there are 4 paired places to put them.

4!*2^4=384

30*384 =11520  possibilities

3840+15360+11520 = 30720 possibilities.

Since this answer agrees with your book it has a fair chance of being correct.

Jun 29, 2019
#5
+109512
+1

ALSO

You have asked far too many questions in a short space of time.

This is an abuse of the forum.

You need to ask a question and learn from the answer before you ask the next question.

You also need to interact with answerers if you want to make any degree of a positive name for yourself here.

Jun 29, 2019
#8
+934
0

Yeah, I was offline for a while because I have a lot of activities during the summer. I'll try to be online more often.

dgfgrafgdfge111  Jun 29, 2019
#6
+109512
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Oh, and if you have an answer provided please explain that, and give it, in your original question.

You should not give us less information than you have for yourself.

You should also explain what you have already done in order to attempt to answer the question yourself.

Jun 29, 2019
#9
+934
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I attempted this question many times but every single time I got different answers. My method was really long, so it would have been very hard to write all that here, including the diagrams. Sorry for the inconvenience.

dgfgrafgdfge111  Jun 29, 2019
#7
0

The number of combinations for n pairs is going to be the (n+1)th fibonacci number times 2n times n factorial

Jun 29, 2019