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

#6**+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

#3**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

#5**+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**+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**+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