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

Arthur tells you, the books are already selected.

Also lulu points out, condition "three particular books are never together" is not very clear, I take it to mean only that all three cannot be together, but two together and one apart is fine.

Now going your way, subtract from total ways, the ways in which all three are together.

Total ways are $12!$ and ways in which three are together are $10!\cdot 3!$ so our answer should be $12! - 3! \cdot 10!$

Answer 2

*THE MISTAKES OF YOUR THINKING * At first, you don't need to select them, they are already selected. Secondary, why you trying to select 3 place from 10 slots, that can't counts the possibilities where 2 of them stay together.
*THE RIGHT WAY TO THINK * at first calculates the all possibilities as $12!$ Then disclude the possibilities , when they all stay together as $ 3!×10!$ Then disclude it from all possibilities . The right answer is $12! -(3!×10! )$

Answer 3

"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 $9!\cdot720=261\,273\,600$.