Some problems deal with Permutations with Restrictions. The best way to illustrate this concept is to contrast it with permutations without restrictions.

Permutations without restrictions

How many ways can the letters $ABCDE$ be distinctly arranged?

Permutations with restrictions

How many ways can the letters $ABCDE$ be distinctly arranged whereby a vowel is not at either end?

How Do We Solve?

The First Case

In the first case, we simply use the "choice" method and say to ourselves well five letters can go in the first spot, four letters in the second, three in the third, two in the fourth, and one in the last.

$$5 \times 4 \times 3 \times 2 \times 1 = 120$$

Imagine there are little horizontal hash marks below each number.

The Second Case

In the second case, we still use the "choice" method but we account for the restriction.

In the first place we cannot have a vowel. I can use $B$, $C$, or $D$. So that's three choices.
In the last place, we also cannot use a vowel. Two letters are remaining of the original three non-vowels. So that's two choices.
For the rest of the places, there are no restrictions, so three choices for the second spot, two choices for the third, and one for the fourth:

$$3 \times 2 \times 3 \times 2 \times 1 = 36$$

Imagine there are little horizontal hash marks below each number.