The Number of Ways to Arrange k Items = k!

A child is playing with four different-colored blocks and is stacking them on top of one another to build a tower. If the child always places the blue-colored block on the bottom, how many different towers can be made?

Incorrect. While there are only three blocks left to arrange, there will be more than three ways to arrange those blocks. [[snippet]]

Correct. If the child always puts the blue block on the bottom, then the number of towers is the number of ways to arrange the remaining three items. Since you are only worried about the three spots on top of the blue block, you can just call those three spots the bottom, middle, and top. $$\displaystyle \stackrel{\mbox{bottom}}{\fbox{3}} { } \stackrel{\mbox{middle}}{\fbox{2}} { } \stackrel{\mbox{top}}{\fbox{1}}$$ The total number of arrangements will be the product of these values. $$\displaystyle 3\cdot 2\cdot 1 = 6$$ You can also use 3! to calculate the number of ways to arrange three blocks. $$\displaystyle 3! = 3\cdot 2\cdot 1 = 6$$ There are six different towers that the child can build.

Incorrect. [[snippet]] There are not 12 different ways to arrange the remaining three blocks. It may help to use __SeBoxes__ to visualize this problem. Since you are only worried about the three spots on top of the blue block, you can just call those three spots the bottom, middle, and top. $$\displaystyle \stackrel{\mbox{bottom}}{\fbox{3}} { } \stackrel{\mbox{middle}}{\fbox{2}} { } \stackrel{\mbox{top}}{\fbox{1}}$$

Incorrect. [[snippet]] The answer 24 is the number of ways to arrange four different items, but you are only counting the number of ways to arrange three different items. $$\displaystyle 24 = 4\cdot 3\cdot 2\cdot 1 = 4! \neq 3!$$

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