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Jan 28, 2014
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Jedithious:

The contestants in a chess tournament were numbered from 1 to 18. When the players were paired for the first game, the sum of the two numbers for each pair was a perfect square. What were the pairings for the first game?


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I remember these from one of my logic puzzle books a couple of years ago. It real easy to do.

You have to make a chart like this. 1/2 chart really but I made the whole thing because you can see it easier.

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

1 x---4---------9---------------------------16
2 -x-----------9------------------------16
3 --x------9-----------------------16
4-----x-9----------------------16
5-------x-------------------16
6----------x-----------16
7------------x---16---------------------------------------25
8----------------x-------------------------------------25
9------------------x-------------------------------25
10---------------------x-----------------------25
11-------------------------x---------------25
12-----------------------------x-------25
13----------------------------------x
14---------------------------------------x
15--------------------------------------------x
16-------------------------------------------------x
17-----------------------------------------------------x
18----------------------------------------------------------x

Then where the row and column intersect put in the sum.

Then erace all numbers that are not perfect squares or just put only those in the first place'

Mark off all matchin numbers because a person cant play with himself -- tho when i do I always win the game (haha)

Notice there is only one way to make a 4 so remove all number on row 1 because you can only use them once the numbers removed are 9 and 16
There is only one in each collom any way so you don't have wory about removing them.
Then make a list to keep track like I did at the bottom.
4 is the only one by it self so 1 and 3 have to be used for this.
16 is the only one on the 2s row so we have to use 2 and 14.


keep track list
4(1,3) 16(2,14) 16(4,12) 16(5,11) 25(7,18) 25(8,17) 25(9,16) 25(10,15) 25(12,13)

These are the answers

This is the first question I answered here.

by Puzzled

PS I saw The Dumb Guy answer and your not dumb if you can do this.

the only difference is the way I did it includes the perfect square of 4.

Puzzled
Jan 28, 2014
 #1
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Hi Jedithious, it is great to see you back again.


"In how many different ways can you arrange 2 blue blocks and 2 red blocks in a straight line?"

I believe the answer is 4P4 / ( 2P2 * 2P2)

there is a nPr button on your calculator or

nPr = n! / (n-r)!
so 4P4 = 4! = 2 x 3 x 4 = 24 Thats how many ways 4 different coloured blocks can be arranged in line.
Then you have to divide by
2P2 because there are 2 red ones 2P2 = 2! / 0! = 2
and you have to divide by 2P2 again because there are 2 red blocks 2P2=2

So I think the answer is 24 / (2*2) = 6

These are the possible permutations RRBB,RBRB,RBBR,BRRB,BRBR,BBRR THAT IS IT!
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Now, lets look at how many ways you can order 4 different books in a more straight forward way.
There are 4 ways to choose the first book
now there are 3 books because there is one on the shelf already.
so there are 3 ways to choose the 2nd book
there are 2 ways to choose the 3rd book and
there is only 1 book left so there is only one way to choose it.
So, there are 4*3*2*1 ways of ordering the 4 books
this is 4!, it is also 4P4.
The P stands for permutations.
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Now you can do the other 2 by yourself I think
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You might find this interesting.
http://www.mathsisfun.com/combinatorics/combinations-permutations.html
Jan 28, 2014
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Jan 28, 2014
Jan 27, 2014
 #2
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Jan 27, 2014

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