[Math] Arranging $12$ different books on a shelf so that $3$ particular books are never together

Question

In how many ways can $12$ different books be arranged on a shelf so that three particular books are never together?

They did take away method as Total ways – always together

But I want to do it like this

first I will select $3$ particular books out of $12$ books then I will arrange those $3$ books in $10$ slots. But I am not getting my answer.

Please help

Thanks

Answer 1

"Never together" means: No two of the three particular books are allowed to be adjacent. If they would have meant "not all three together" they would have said so.

You can arrange the $9$ ordinary books in $9!$ ways, creating $10$ slots where one of the three particular books may be placed. There are $10\cdot9\cdot 8$ ways to choose different slots for these three books.

The total number of admissible arrangements therefore is $9!\cdot720=261\,273\,600$.