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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

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

  2. Empty
    • one year ago
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    Are parts considered books

  3. ParthKohli
    • one year ago
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    Is it D?

  4. mathmath333
    • one year ago
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    Yes answer is given \(D.)\)

  5. ParthKohli
    • one year ago
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    OK. It's because it is D.

  6. mathmath333
    • one year ago
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    Example \(\text{ Book C part 1}\) and \(\text{ Book C part 2}\) are two different books

  7. Empty
    • one year ago
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    \[\frac{(1*20+2*12+3*5+1*1+1*1+1*1)!}{20! 12!^25!^3}\]

  8. ParthKohli
    • one year ago
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    yeah, really needs no explanation.

  9. mathmath333
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    Sorry, that's six distinct letters.

  13. Empty
    • one year ago
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    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.

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