A community for students.
Here's the question you clicked on:
 0 viewing
mathmath333
 one year ago
A library has 20 copies of book A ,12 copies each of
book B part 1 and part 2 , 5 copies of book C part 1
,part 2 and part 3 and single copy of book D, book E,
book F. In how many ways can these book be distributed
mathmath333
 one year ago
A library has 20 copies of book A ,12 copies each of book B part 1 and part 2 , 5 copies of book C part 1 ,part 2 and part 3 and single copy of book D, book E, book F. In how many ways can these book be distributed

This Question is Closed

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0\(\large \color{black}{\begin{align} & \normalsize \text{ A library has 20 copies of book A ,12 copies each of}\hspace{.33em}\\~\\ & \normalsize \text{ book B part 1 and part 2 , 5 copies of book C part 1 }\hspace{.33em}\\~\\ & \normalsize \text{ ,part 2 and part 3 and single copy of book D, book E,}\hspace{.33em}\\~\\ & \normalsize \text{ book F. In how many ways can these book be distributed}\hspace{.33em}\\~\\ & a.)\ \dfrac{62!}{20!12!5!} \hspace{.33em}\\~\\ & b.)\ 62! \hspace{.33em}\\~\\ & c.)\ \dfrac{62!}{(37)^{3}} \hspace{.33em}\\~\\ & d.)\ \dfrac{62!}{20!(12!)^{2}(5!)^{3}} \hspace{.33em}\\~\\ \end{align}}\)

Empty
 one year ago
Best ResponseYou've already chosen the best response.3Are parts considered books

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0Yes answer is given \(D.)\)

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2OK. It's because it is D.

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0Example \(\text{ Book C part 1}\) and \(\text{ Book C part 2}\) are two different books

Empty
 one year ago
Best ResponseYou've already chosen the best response.3\[\frac{(1*20+2*12+3*5+1*1+1*1+1*1)!}{20! 12!^25!^3}\]

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2yeah, really needs no explanation.

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0i didnt understand the denominator \(\dfrac{(1*20+2*12+3*5+1*1+1*1+1*1)!}{\color{red}{20! 12!^25!^3}}\)

Empty
 one year ago
Best ResponseYou've already chosen the best response.3There are 20! ways to arrange the A books. But since these are indistinguishable, we divide this quantity out of the total since 62! treats each book as if they were independent when they're not.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2If I have these letters and I want to find the number of arrangements... ``` AAABBC ``` I will first find the number of arrangements of five distinct letters (5!) then divide by 3!*2! because A repeats thrice and B repeats twice.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2Sorry, that's six distinct letters.

Empty
 one year ago
Best ResponseYou've already chosen the best response.3It's probably better to think through this question with a small size, well @ParthKohli is on it and I'm not in India so it's 4AM here later lol.
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.