Question

Amanda has four plastic shapes, a circle, a square, a triangle, and a pentagon, In how many ways can she lineup the four shapes if the circle cannot be next to the square

Do NOT have an explanation that isn't just writing all the combinations of shapes.

Answer 1

The number of ways Amanda can lineup the four shapes when the circle cannot be next to the square is 6.

Step-by-step explanation:

The number of combinations of lining up the four shapes, a circle, a square, a triangle, and a pentagon, can be found using the concept of permutations. In this case, since the circle cannot be next to the square, we can treat the circle and the square as a single entity. So, we have 3 entities: (circle and square), triangle, and pentagon. The number of permutations of these entities is given by 3! (3-factorial), which is equal to 3x2x1 = 6.

Answer 2

Answer: 12

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Step-by-step explanation:

Let's say we have the following code names for each shape

• C = circle
• S = square
• T = triangle
• P = pentagon

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.