Find the number of distinct line ups such that A,B,C are not adjacent?

Question

question:

10 peoples including A,B,C are waiting in a line.How many distinct line ups are there such that A,B,C are not adjacent?(assumption: A,B,C may be in any order as long as all three are adjacent) My attempts:

(i) first i put A,B,C in 8 gaps formed by 7 peoples (2 in sides and 6 in between)by $^8C_3$ and arranged rest 7 peoples by $7!$ and got answer as $^8C_3$$7!=282240$ ways, but answer is coming '3386880'i.e,(10!-8!.3!)

(i think my answer above is wrong because i only calculated those cases where no two peoples are consecutive and left the case where two are consecutive and one is not)

(ii)then,i tried again using $(total-together=not together)$ so, i treated A,B,C as single entity and calculated $(10!-8!=3588480)$(i didn't multiplied 3! to 8! because by assumption given in the question order of A,B,C isn't relevant )

(iii)then, I again tried using formula $(total - together=not together)$ by calculating total cases as$[10!-3!7!]$ and subtracted cases treating A,B,C as single entity i.e, $8!$ and got answer as $(10!-3!7!-8!)=3558240 $ again got answer wrong

i know i'm doing conceptual mistakes ....please enlighten me.

and tell how to approach further....thank you.

Answer 1

In your second approach (ii), you should have multiplied $8!$ by $3!$ so the answer should be $10!-8!\cdot3!$. Actually I don't really understand the assumption you have written since it says "as long as they are adjacent" but we are looking for the cases where all three are not adjacent. But I think it is made in order to emphasize two of them may be adjacent as long as all three are not adjacent. Logically, when you change the places of $A,B$ and $C$ inside when they are adjacent, each permutation of them will give a distinct line therefore the answer should be $10!-8!\cdot3!$.

Answer 2

There are $3!$ ways for $A,B,C$ to be ordered. Now, given a specific ordering of $A,B,C$ (say, $CAB$), treat this ordering of $A,B,C$ as just $1$ person, and you have $8!$ ways to arrange $7$ people and that specific ordering of $A,B,C$. So the total number of ways that $A,B,C$ can end up adjacent to each other is given by the product of these two: 3!8!. Thus, the number of ways to arrange these people so that $A,B,C$ are not adjacent is $10! - 3!8!$.