Total number of permutations are $6!$ and there are $4!$ ways where the word $UP$ shows up and $2!$ ways where the word $LOCK$ shows up. Also, there are two arrangements where both of the the words show up(LOCKUP and UPLOCK). So according to me the final answer must be
$6!-4!-2!-2$
Am I wrong in this answer?
On
The number of words formed by permuting the letters L,O,C,K,U,P such that neither 'LOCK' nor 'UP' appears in any such arrangement
The total number of arrangements with no restriction is $6!$
The number of arrangements where LOCK appears is $3!$ because we take lock as one unit. Therefore, we are left with LOCK,U,P
The number of arrangements where up appears would be $5!$ because we Take UP as one unit and we are left with four other letters. L, O, C, K, UP
The number of arrangements where LOCK and UP appear is $2!$ by taking UP as one unit and LOCK as one unit.
Final answer: $6!-(5!+3!-2!)$
You were subtracting everything from each other, so let me explain to you what we need to do exactly.
Using inclusion and exclusion principle... In your case, you will need to use this rule
$|A \cup B| = |A|+|B|-|A \cap B|$
Then the answer would be $|X| - |A \cup B|$ where X is the total number of arrangements with no restriction, A represents the number of arrangements where LOCK appears, and B represents the number of arrangements where UP appears, then finally $|A \cap B|$ where LOCK and UP appear together.
MY ANSWER IS 6! - 5! - 3! +2
THE TOTAL FORMS ARE 6! THE FORMS THAT "UP" APPEARS ARE 5! THE FORMS THAT "LOCK" APPEARS ARE 3! AND THE FORMS BOTH APPEARS IS 2 THEN WE COUNT IT TWICE AND MUST ADD UP IT