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How many ways can $5$ books be arranged on a shelf if $2$ of the books must remain together?

I have $5$ books $A,B,C,D,E $ and spots as $$\underline{A} \ \underline{B} \ \underline{C} \ \underline{D} \ \underline{E}$$

These can be arranged in $5!$ ways. So suppose that $A,B$ must remain together. Can I treat $AB$ as a single element and consider $$\underline{AB} \ \underline{C} \ \underline{D} \ \underline{E}$$ even though this has $5$ books but $4$ spots? These can be arranged as $$\underline{AB} \ \underline{C} \ \underline{D} \ \underline{E} \\ \underline{C} \ \underline{AB} \ \underline{D} \ \underline{E} \\ \vdots \\ \underline{C} \ \underline{D} \ \underline{E} \ \underline{AB} $$ so I have $4$ different arrangements that can be ordered in $4!$ ways so the total would be $5! - 4\cdot(4!)= 24$?

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    $\begingroup$ It can either be $AB$ or $BA$ $\endgroup$
    – Math Lover
    Nov 25 at 9:56
  • $\begingroup$ Also $5! - 4 \cdot 4!$ is nothing but $4!$ as $AB$ together gives you $4$ books (one of them being a combined book) $\endgroup$
    – Math Lover
    Nov 25 at 9:57
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    $\begingroup$ If you treat AB as one item then you get $4!$ possibilities, but BA counts too so $2 \times 4!$ combined $\endgroup$
    – Henry
    Nov 25 at 10:03

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