combinatorics - Within the number of different ways to order in line the letters of the word LOVEPONIXO, in how many ways there are no two O's one after another?

Within the number of different ways to order in line the letters of the word LOVEPONIXO, in how many ways there are no two O's one after another?

I was thinking - there are $\frac{10!}{3!}$ different ways to order the letters. Then I want to subtract the number of ways that include two O's one after another: $2!\cdot\frac{9!}{3!}$
And then I want to subtract the number of ways that include three O's on after another (some of them are included in the last calculation) : $3!\cdot\frac{8!}{3!}$

Then I get a final answer: $\frac{10!}{3!}-2!\cdot\frac{9!}{3!}-3!\cdot\frac{8!}{3!} = 8\cdot\frac{9!}{3!}-8!$

I have a strong feeling that I did something wrong though... Any directions?

Answer

A simpler method of doing the problem is to set aside the three O's for the moment. There are $7!$ ways of arranging the other seven letters since they are distinct. For any given arrangement, we now have eight gaps, six between successive letters and the two ends, in which to place the three O's. We can place the O's in these gaps in $\binom{8}{3}$ ways. Hence, the number of permutations in which no two O's are consecutive is
$$7! \cdot \binom{8}{3}$$

The method you attempted can be made to work. As you observed, there are $$\frac{10!}{3!}$$ distinguishable arrangements of the letters. From these, we must exclude those in which at least two O's are consecutive.

Suppose two O's are consecutive. We treat them as one letter, say $\cal{O}$. Thus, we have nine objects to arrange. Since they are distinct, they can be arranged in $9!$ ways.

However, when we subtract those arrangements in which at least two O's are consecutive, we remove those arrangements in which three O's are consecutive twice. Therefore, we must add them back in. If we treat the three O's as one object, we have eight distinct objects to arrange, which can be done in $8!$ ways.

Thus, by the Inclusion-Exclusion Principle, the number of distinct permutations of LOVEPONIXO in which no two O's are consecutive is
$$\frac{10!}{3!} - 9! + 8!$$
As you can verify by direct calculation, this agrees with the answer given above.