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. 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 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 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:
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How many different selections of 5 books can be made from 12 different books if, Two particular books are always selected? Total number of books = 12 Number of books to be selected = 5 Given Two books are always selected. Remaining number of books to be selected = 3 The number of ways of selecting the remaining 3 books from the remaining 10 books = 10C3 = `(10!)/(3! xx (10 - 3)!)` = `(10!)/(3! xx 7!)` = `(10 xx 9 xx 8 xx 7!)/(3! xx 7!)` = `(10 xx 9 xx 8)/(3!)` = `(10 xx 9 xx 8)/(3 xx 2 xx 1)` = 5 × 3 × 8 = 120 ways Is there an error in this question or solution? Page 2How many different selections of 5 books can be made from 12 different books if, Two particular books are never selected? Two particular books are never selected. Since two books are never selected. The total number of books is 10. ∴ The number of ways of selecting 5 books from 10 books = 10C5 = `(10!)/(5! xx (10 - 5)!)` = `(10!)/(5! xx 5!)` = `(10 xx 9 xx 8 xx 7 xx 6 xx 5!)/(5! xx 5!)` = `(10 xx 9 xx 8 xx 7 xx 6)/(5!)` = `(10 xx 9 xx 8 xx 7 xx 6)/(5 xx 4 xx 3 xx 2 xx 1)` = 2 × 9 × 2 × 7 = 252 ways Is there an error in this question or solution? |
How many solutions of 5 books can be made from 12 different books if two particular books are always selected?
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