SOLUTION: On my bookshelf I have 5 books of poetry, 15 graphic novels, 8 biographies and 3 self-help books. I plan to take 7 books with me on my vacation. a)In how many different ways ca

Algebra ->  Permutations -> SOLUTION: On my bookshelf I have 5 books of poetry, 15 graphic novels, 8 biographies and 3 self-help books. I plan to take 7 books with me on my vacation. a)In how many different ways ca      Log On


   



Question 1169264: On my bookshelf I have 5 books of poetry, 15 graphic novels, 8 biographies and 3 self-help books. I plan to take 7 books with me on my vacation.
a)In how many different ways can I take 7 books with me on my vacation?

b)In how many different ways can my 7 books consist of 2 books of poetry, 3 graphic novels and 2 self-help books?
c)In how many different ways can my 7 books consist of exactly 3 graphic novels?
d)In how many different ways can my 7 books consist of at least 6 biographies?

Answer by ikleyn(52778) About Me  (Show Source):
You can put this solution on YOUR website!
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On my bookshelf I have 5 books of poetry, 15 graphic novels, 8 biographies and 3 self-help books.
I plan to take 7 books with me on my vacation.
a)In how many different ways can I take 7 books with me on my vacation?
b)In how many different ways can my 7 books consist of 2 books of poetry, 3 graphic novels and 2 self-help books?
c)In how many different ways can my 7 books consist of exactly 3 graphic novels?
d)In how many different ways can my 7 books consist of at least 6 biographies?
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(b)  In how many different ways can my 7 books consist of 2 books of poetry, 3 graphic novels and 2 self-help books?


     You can select 2 books of poetry from  5 books of poetry by C%5B5%5D%5E2 = %285%2A4%29%2F%281%2A2%29 = 10 ways.


     You can select 3 graphic novels  from 15 graphic novels  by C%5B15%5D%5E3 = %2815%2A14%2A13%29%2F%281%2A2%2A3%29 = 455 ways.


     You can select 2 self-help books  from 3 self-help books  by C%5B3%5D%5E2 = %283%2A2%29%2F%281%2A2%29 = 3 ways.


     Since these selections are INDEPENDENT in separate categories, the total number of choices is the product 10*455*3 = 13650,

        according to the foundamental counting principle.

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On Combinations,  see introductory lessons
    - Introduction to Combinations
    - PROOF of the formula on the number of Combinations
    - Problems on Combinations
    - Fundamental counting principle problems
    - OVERVIEW of lessons on Permutations and Combinations
in this site.

Also,  you have this free of charge online textbook in ALGEBRA-II in this site
    - ALGEBRA-II - YOUR ONLINE TEXTBOOK.

The referred lessons are the part of this online textbook under the topic  "Combinatorics: Combinations and permutations".


Save the link to this textbook together with its description

Free of charge online textbook in ALGEBRA-II
https://www.algebra.com/algebra/homework/complex/ALGEBRA-II-YOUR-ONLINE-TEXTBOOK.lesson

into your archive and use when it is needed.