## 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

1. mathmath333

\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}}

2. Empty

Are parts considered books

3. ParthKohli

Is it D?

4. mathmath333

Yes answer is given $$D.)$$

5. ParthKohli

OK. It's because it is D.

6. mathmath333

Example $$\text{ Book C part 1}$$ and $$\text{ Book C part 2}$$ are two different books

7. Empty

$\frac{(1*20+2*12+3*5+1*1+1*1+1*1)!}{20! 12!^25!^3}$

8. ParthKohli

yeah, really needs no explanation.

9. mathmath333

i didnt understand the denominator $$\dfrac{(1*20+2*12+3*5+1*1+1*1+1*1)!}{\color{red}{20! 12!^25!^3}}$$

10. Empty

There 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.

11. ParthKohli

If 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.

12. ParthKohli

Sorry, that's six distinct letters.

13. Empty

It'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.