How many ways can 5 books be selected out of 10 books if two specific books are never selected?

Text Solution

Solution : (i) Since, two particular books are always selected. It means that 6y-2=4 books are selected out of the remaining 11-2=9 books. <br> `therefore`Required number of ways`=.^(9)C_(4)=(9*8*7*6)/(1*2*3*4)=126`. <br> (ii) Since, two particular books are never selected, it means that 6 books are selected out of the remaining 11-2=9 books. <br> `therefore` Required number of ways=`.^(9)C_(6)` <br> `=.^(9)C_(3)=(9*8*7)/(1*2*3)=84`.

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

What matters is which books you borrow, not how many books you borrow. Borrowing Apostol's Calculus and Munkres' Topology is different from borrowing Herstein's Algebra and Rudin's Principles of Mathematical Analysis.

There are $\binom{5}{k}$ ways for you to borrow exactly $k$ of the five books. Since you can borrow from $0$ to $5$ books, the number of different selections of books you could borrow is $$\sum_{k = 0}^{5} \binom{5}{k} = \binom{5}{0} + \binom{5}{1} + \binom{5}{2} + \binom{5}{3} + \binom{5}{4} + \binom{5}{5} = 32$$ As Parcly Taxel observed, with each selection, you either choose to borrow a particular text or not borrow it. Since there are two choices for each of the five books, there are $2^5 = 32$ possible selections of books you could make.

The two approaches are related by the Binomial Theorem. $$(x + y)^n = \sum_{k = 0}^{n} \binom{n}{k}x^{n - k}y^k$$ If we let $n = 5$, $x = 1$, and $y = 1$, we obtain $$2^5 = (1 + 1)^5 = \sum_{k = 0}^{5} \binom{5}{k}1^{n - k}1^k = \sum_{k = 0}^{5} \binom{5}{k}$$