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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 $5! - 4\cdot[4!]= 24$?
asked Nov 25, 2021 at 9:53
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As mentioned in the comments, you must also consider the order of the two books that remain together.
Place the two books that must remain together in a box. Then we have four objects to arrange, the box and the other three books. The four objects can be arranged in $4!$ orders. The two books within the box can be arranged in $2!$ orders. Hence, the number of possible arrangements of five books if two of the books must remain together is $4!2!$.
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Here are the five books: Let's use slots like we did with
the license plates:
We'll fill each slot -- one at a time... Then we can use the counting principle!
The first slot:
We have all 5 books to choose from to fill this slot.
Let's say we put book C there...
Now, we only have 4 books that can go here...
How many books are left for this slot?
See it?
Whoa, dude! That's 5!
So, there are 120 ways to arrange five books on a bookshelf.
[Aren't you glad I didn't make you draw them out?]
Was the answer to our 3-book problem really 3! ?
Yep!
Will this always work?
TRY IT:
How many ways can eight books be arranged on a bookshelf? [reason it out with slots]
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Updated On: 27-06-2022
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