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There are 16 benches in a park and 9 big bushes.

 

The benches are set out by 4x4, so a huge square and thee bushes are placed in the centre of every tiny (4) squares.

Also displayed as,

 

 

There are 4 spies in the park who are making secret phone calls, so they don't want be seen by each other!

 

How many different possibilities are there for each Spy to sit on a bench without being seen by each other?

(Can you use permutation to solve this?)

 

Please show workings so I can understand thanks!

 

PS:Even my brother found this hard!

MathsGod1  May 10, 2015

Best Answer 

 #6
avatar+78643 
+15

I get the same answer as Alan....but with a slightly different explanation...

The first spy can sit in any of the first four "vertical" benches . The next spy has three vertical benches to choose from. The third spy has two and the last one has one. That's 4 x 3 x 2 x 1 = 4! ways = 24 ways. But, once seated, we can also arrange these vertical columns in 4! ways.....so....4! x 4!  = 24 x 24 = 576.

 

 

  

CPhill  May 11, 2015
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8+0 Answers

 #3
avatar+91038 
0

Yes I think this is quite hard M.G.  

It looks like a good puzzle.  I will add it to our puzzles thread   

Melody  May 10, 2015
 #4
avatar+3 
0

4! x 4! = 576 

sean1w0ii  May 10, 2015
 #5
avatar+26329 
+15

They must all sit in different rows and different columns.

 

I think this means there are 16 places for the first spy, 9 places left for the second spy, 4 places left for the third spy and just 1 left for the fourth spy.  So I'm going with 16*9*4*1 = 576 possibilities.

.

Alan  May 11, 2015
 #6
avatar+78643 
+15
Best Answer

I get the same answer as Alan....but with a slightly different explanation...

The first spy can sit in any of the first four "vertical" benches . The next spy has three vertical benches to choose from. The third spy has two and the last one has one. That's 4 x 3 x 2 x 1 = 4! ways = 24 ways. But, once seated, we can also arrange these vertical columns in 4! ways.....so....4! x 4!  = 24 x 24 = 576.

 

 

  

CPhill  May 11, 2015
 #7
avatar+91038 
+5

Okay I get the same answer too but my explanation is a little different again.

I'm going to put spy1 in row 1 that is 4 ways

spy2 i row 2 that is 3 ways

spy3 in row 3 that is 2 ways

spy 4 in row 4 that is 1 way       so so far I have  4*3*2*1 = 4!   Just like CPhill.

 

Now there are 4P4 or 4! permutations for the order that the spys can be chosen in.

So the number of ways is 4!*4! = 576       

Melody  May 11, 2015
 #8
avatar+4663 
+5

Wow you made that look a lot easier...

 

MathsGod1  May 11, 2015
 #9
avatar+91038 
0

Everything is easy when you know how and you have the right tools :))

Melody  May 11, 2015
 #10
avatar+4663 
+5

MathsGod1  May 11, 2015

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