How many distinct 4-letter code words can be made using the letters in the word “SOONERS”?

Question

The available answer choices are: (A) 144 (B) 188 (C) 216 (D) 270 (E) 324

I do not know how to confront this problem; I assume it had to do with permutations and combinations though, since it is about arrangements.

Answer 1

Split this up into cases.

Case 1: Two "S"s and two "O"s

There are only six distinct combos: "SSOO", "OOSS", "OSOS", "SOSO", "SOOS", and "OSSO". This also comes from $4!$ possible combinations divided by $2!$ duplicate "O"s and $2!$ duplicate "S"s, so $\frac{4!}{2! 2!} = 6$

Case 2: Two "O"s

If there are two "O"s, there can only be one "S" or it falls into the above category. Thus you have $\frac{_4C_2 * 4!}{2} = 72$ possible words. $_4C_2$ comes from choosing two remaining letters, $4!$ is how many different ways you can arrange them, and $\frac{1}{2}$ comes from the fact that for every arrangement of both "O"s, there is another one with the "O"s switched which is nondistinct from the first arrangement.

Case 3: Two "S"s

The argument for this is identical to the previous one, so $72$ possible combos.

Case 4: No duplicate letters

The possible letters to choose from are SONER, so it will be $_5C_4$. There are $4!$ possible ways to arrange this, so $_5C_4 * 4! = 120$ distinct combos.

Now add all cases together and you have $270$ possible combinations (if I did this right)