Each square in a 2 ∗ 2 table is colored either black or white. How many different colorings of the table are there?
Note- Rotations and turnings are allowed after colouring once.
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i got 16 .
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They're not all different if rotations and turnings are allowed after coloring once....
think too much of the problem. huhu. made it so complicated (with regards to the solution) hahaha
I think you need to choose better wording. When you say rotations and turnings are allowed after colouring once, it makes it seem that the answer should be 16.
In other words, if you paint 1 box, it should be able to be done 4 ways because rotations are allowed.
If you changed it to say rotations do not count as a separate coloring, it would be very clear what you were asking
Looks like it is 4 ! / 4
This is because each coloring can be transformed in 4 ways by turning it 2 π radians everytime.
Good! Perhaps it is a fine solution
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Doesn't each box get painted either black or white? Since rotations and turning is allowed, it must be implying that the 4 boxes are different. So, each box has 2 cases - white or black. So, 2x2x2x2 = 16.
Please tell me where I am going wrong.
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Sorry if I am sounding wrong, but- Just try it on your own. You'd get it. And if you don't , pl look at Aditya's solution
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@Krishna Ar – 16 is correct as per law of combinatronics
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@Shashank Chauhan – yes 16 is correct if the table is static, but he said in the problem " Rotations and turnings are allowed after coloring once " . So if we look at the black as '1' and the white as '0' the combination of '0001' after rotation and turning may be '0100' or '0010' or '1000' then a 4 combinations for the same case after rotation have the same color . Do the same for the other combinations '0011','0110','0111'. Note that '0000' '1111' not affected by rotation . the result will be 6 different color
@Shashank Chauhan – Nope.
1.There is no law applicable for such problems. These type of problems are known as casework problems.
That doesn't work because there are 16 ways to color it without rotations
The first we hear these 2 statements
If you paint 0 boxes , it can be done in 1 way
it looks really odd , so how can you explain this to a small kid in your way?
How many wikis have you posted till now - very eager to read all
and you too @Aditya Raut - how would you explain?
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By splitting into cases,
I f y o u p a i n t 0 b o x e s , it can be done in 1 way.
I f y o u p a i n t 1 b o x , it also can be done in only 1 way.
I f y o u p a i n t 2 b o x e s , that has 2 ways to work with, 2 cases, if you paint adjacent ones then a way and if you paint diagonally opposite then one way - 2 ways
I f y o u p a i n t 3 b o x e s , it's same as not painting 1 box , same as case 2 hence 1 way.
I f y o u p a i n t a l l 4 b o x e s ,1 way.
Total 1 + 1 + 2 + 1 + 1 = 6