Question

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

Answer 1

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

Answer 2

Are parts considered books

Answer 3

Is it D?

Answer 4

Yes answer is given \(D.)\)

Answer 5

OK. It's because it is D.

Answer 6

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

Answer 7

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

Answer 8

yeah, really needs no explanation.

Answer 9

i didnt understand the denominator

\(\dfrac{(1*20+2*12+3*5+1*1+1*1+1*1)!}{\color{red}{20! 12!^25!^3}}\)

Answer 10

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.

Answer 11

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.

Answer 12

Sorry, that's six distinct letters.

Answer 13

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.