In how many ways can 6 different books be arranged on a shelf of 2 books are always together

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)

Page 2

Now, we're going to learn how to count and arrange. (As if just learning to count wasn't exciting enough!)

How many ways can we arrange three books on a bookshelf?

Here are the books:

Well, there's one arrangement.

Let's pound out the others:

That's all of them... There are 6 ways to arrange three books on a bookshelf.

What about five books?

Dang! I don't want to have to draw it all out!

Let's FIGURE it out instead.

Page 3

* For this one, order does NOT matter!

We did this problem before:

If we have 8 books, how many ways can we arrange 3 on a
bookshelf?

We figured it out with slots:

But, using the formula gave us the same thing:

Here's a different question for you:

If we have 8 books and we want to take 3 on vacation with us, how
many ways can we do it?

What's the difference between these problems?

ORDER DOESN'T MATTER!

In the first problem, we were arranging the 3 books on a shelf... and in the second problem, we're just tossing the 3 books in a suitcase.

So, if order doesn't matter, we'll just divide it out!

Arranging the 3 books is 3!

Page 4

Grab a calculator! I'm going to teach you about a new button.

Look for it... It will either be

(It's probably above one of the other buttons.)

Find it?

It's called a factorial.

Here's an example:

(No, this isn't just an excited 5.)

Here's what it means:

Check it by multiplying it out the long way, then try the button.

Here are some others:

Page 5

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Suppose you have six different books on a shelf with labels $A, B, C, D, E,$ and $F$. In how many different ways can you arrange the books on the shelf if books $A, B,$ and $C$ are grouped together?

(e.g. $EBACFD$ is an acceptable arrangement, but $EBAFDC$ is not an acceptable arrangement)

While trying to solve it I get $120$, by $6\times5\times2\times1\times2\times1$. Although the correct answer is $144$. Can anyone correct me on where I am going wrong?

Answer

Verified

Hint: To solve the above question we will use the term factorial (!). Actually factorial is a function that multiplies a number by every number below it. For example, we have $5!$ , then we can write it as $5!=5\times 4\times 3\times 2\times 1$. This factorial function is used among other things, to find the number of ways “n” objects can be arranged. This factorial term is used in many mathematical formulas such as permutation and combination.

Complete step by step solution:

Now in the above question we have to find the number of ways by which six books can be arranged on a self.In mathematics, there are n! ways of arranging n distinct objects into an ordered sequence. The factorial n! gives the number of ways in which n objects can be permuted. For example, $2$ factorial is $2!=2\times 1$ , it means there are two different ways to arrange the numbers $1$ through $2$ which are$\left\{ 1,2 \right\},\left\{ 2,1 \right\}$.Now to arrange six books on a shelf, that means we have six spots to put them in.Our first book has six spots it can be in, and then our second book has $5$ , since one is taken by the first book. Then the next will have $4$, and then $3$ , then$2$ , then $1$. Thus we can write it as:$\Rightarrow 6!=6\times 5\times 4\times 3\times 2\times 1=720$

Hence we have $720$ ways to arrange the books on shelf.

Note: The above question is simple because here is an arranged number on a shelf. But sometimes we have to arrange the letters in a circular form, so for the circular form the number of arrangements of elements is$\left( n-1 \right)!$.