# Check my work on permutations of letters

In how many arrangements can the word "CORONA" be arranged in only 4 letters with no restriction?

I divided into 2 cases:

Case 1: No doubles, hence we have 1 C, 1 O, 1 R, 1 N and 1 A to arrange into 4 spaces, so 5P4, in other words 5!.

Case 2: one double, we have a double O, to arrange into 4 spaces with repetition, so $$\frac{4P2}{2!}$$ and the rest of the letters C, R, N and A, to be arranged in the 2 remaining spaces, hence $$\frac{4P2}{2!}\cdot 4P2$$

So the final answer will be adding both: $$\frac{4P2}{2!}\cdot 4P2+5!=192$$

Is this correct?

In how many arrangements can the word "CORONA" be arranged in only 4 letters with no restriction?

Yes your work is both accurate and valid. However, I found it a bit confusing.

I would have worded it this way.

$$\underline{\text{case 1: 0 O's}}$$
4! ways to permute the C, R, N, A.
4!

$$\underline{\text{case 2: 1 O}}$$
$$\binom{4}{3} = 4$$ ways to choose the other three letters.
Then, 4! ways to permute the O with the other 3 letters.
$$4 \times 4!$$

$$\underline{\text{case 3: 2 O's}}$$
$$\binom{4}{2} = 6$$ ways to choose the other two letters.
Then, $$4 \times 3 = 12$$ ways to place the other two letters.
$$6 \times 12$$

$$(4!) + (4 \times 4!) + (6 \times 12).$$