Video solution:

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Written solution:

Imagine adding an extra coat hook with a coat on it after the last coat. Now, there will be $36$ coat hooks, and a repeating pattern, where each block of the pattern has a bunch of empty hooks followed by a coat. Let $b$ be the number of items in each block. Let $d$ be the number of blocks. Note that $d$ is exactly one more than the number of coats, because we added an extra coat at the end.

We then have $bd = 36.$

The constraint on $b$ is that $b \geq 2$ because each block has at least one empty hook, and ends with a coat.

The constraint on $d$ is that $d \geq 2$ because there was at least one coat before, and we added one extra coat at the end.

So, we just need to find out how many ways there are to factorize $36$ into the product of two integers that are at least $2.$

The number of ways to factorize $36$ into the product of two positive integers is exactly equal to the number of factors of $36.$ There is a formula for that: since $36 = 2^2 \times 3^2$, the number of factors of $36$ is $$(2+1)(2+1) = 9.$$

Out of these factorizations, exactly two are disqualified: $1 \times 36$ and $36 \times 1.$ So, the answer is $9 - 2 = 7,$ which is choice D.