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Explanation:
Let's say we have the following code names for each shape
We have 4! = 4*3*2*1 = 24 ways to arrange them without any restrictions
Some permutations have C and S together, while others do not.
Example of them together: CSTP
Example of them separated: CTSP
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Let's say Amanda used a rubberband or glue to connect the circle to the square. This way the two shapes would always be together. They combine to form a new shape of sorts.
Let's call this new shape "polygon" and define it as
L = polygon
The codename L is just a placeholder for CS or SC.
So instead of the set of these codenames {C,S,T,P}, we have this reduced set {L,T,P}
No matter how we arrange the items in {L,T,P}, the circle and square are always going to be together. Again, anywhere you see an L, you replace it with CS or SC.
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There are 3 letters in {L,T,P} so there are 3! = 3*2*1 = 6 ways to arrange those 3 letters. There are 2 ways to arrange S and C within any of those 6 arrangements mentioned earlier.
So there are 2*6 = 12 different ways to arrange the four shapes such that S and C are together somehow.
Recall earlier we found 24 ways total to arrange the shapes without restriction. This must mean there are 24 - 12 = 12 permutations such that S and C are not together. This is the final answer we're after.
