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mezya [45]
3 years ago
5

A librarian has 4 identical copies of Hamlet, 3 identical copies of Macbeth, 2 identical copies of Romeo and Juliet, and one cop

y of Midsummer’s Night Dream. In how many distinct arrangements can these ten books be put in order on a shelf?
Mathematics
1 answer:
lesantik [10]3 years ago
4 0

Answer:

The number of distinct arrangements is <em>12600</em><em>.</em>

Step-by-step explanation:

This is a permutation type of question and therefore the number of distinguishable permutations is:

n!/(n₁! n₂! n₃! ... nₓ!)

where

  • n₁, n₂, n₃ ... is the number of arrangements for each object
  • n is the number of objects
  • nₓ is the number of arrangements for the last object

In this case

  • n₁ is the identical copies of Hamlet
  • n₂ is the identical copies of Macbeth
  • n₃ is the identical copies of Romeo and Juliet
  • nₓ = n₄ is the one copy of Midsummer's Night Dream

Therefore,

<em>Number of distinct arrangements =  10!/(4! × 3! × 2! × 1!)</em>

<em>                                                         = </em><em>12600 ways</em>

<em />

Thus, the number of distinct arrangements is <em>12600</em><em>.</em>

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A^{-1} = \frac{1}{3493}  \begin{pmatrix} 210 && -59 && -12 \\ 84 && 176 && 95 \\ 77 && -5 && 295 \end{pmatrix}

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Step-by-step explanation:

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A^{-1} = \frac{1}{3493}  \begin{pmatrix} 210 && -59 && -12 \\ 84 && 176 && 95 \\ 77 && -5 && 295 \end{pmatrix}

Then when you multiply with the vector solution

A^{-1} b =  \frac{1}{3493}  \begin{pmatrix} 210 && -59 && -12 \\ 84 && 176 && 95 \\ 77 && -5 && 295 \end{pmatrix} * \begin{pmatrix} 2400 \\ 3500 \\ 4000 \end{pmatrix} = \frac{1}{3493}  \begin{pmatrix} 249500 \\ 1197600 \\ 1347300 \end{pmatrix}\\\\\\= \begin{pmatrix} \frac{500}{7} \\\\ \frac{2400}{7} \\\\ \frac{2700}{7} \end{pmatrix}

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Answer:

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