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There are 6 English books, 4 Science books, 7 magazines, and 3 Mathematics books. In how many ways can you arrange the shelf if: a) English and Science books are indistinct? b) English books should be together?

Pls someone help me on this one. Thank you!

EDIT: I actually have an initial answer. For a, $(10!)/(6! 4!)$ ways for the English and Science books, then multiply to $10!$ (ways for the others)? Is this correct? I'm actually not sure if I understand the restriction in a correctly.

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    $\begingroup$ Have you seen similar questions on the site? By similar, I refer to counting the arrangement of objects of different types. Look at them, and reference any that you find similar in your post above. $\endgroup$ Jan 19 '19 at 9:24
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    $\begingroup$ Please read the related questions at right, then make an attempt at solving the problem yourself. $\endgroup$ Jan 19 '19 at 10:45
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In how many ways can $6$ English books, $4$ science books, $7$ magazines, and $3$ mathematics books be arranged on a shelf if English books are indistinguishable and science books are indistinguishable?

We have a total of $6 + 4 + 7 + 3 = 20$ books. Choose six of the $20$ positions for the English books and four of the remaining $14$ positions for the science books. The remaining ten positions can be filled with books and magazines in $10!$ ways.

$$\binom{20}{6}\binom{14}{4}10! = \frac{20!}{6!14!} \cdot \binom{14!}{4!10!} \cdot 10! = \frac{20!}{6!4!}$$

In your attempt, you did not take into account the total number of positions on the shelf.

In how many ways can $6$ English books, $4$ science books, $7$ magazines, and $3$ mathematics books be arranged on a shelf if English books should be together?

If all the books are intended to be distinct (switching the order of the questions would have made this clearer), treat the English books as a single object, so we have $1 + 4 + 7 + 3 = 15$ objects to arrange. Then multiply by the number of ways of arranging the six English books within the block of English books.

If we are still supposed to treat the English books as being indistinguishable and the science books as being indistinguishable, choose six of the $15$ positions for the science books, one of the remaining $8$ positions for the block of English books, then arrange the magazines and mathematics books in the remaining positions.

I believe the first of these two interpretations is intended, but I would have reversed the order of the questions to make that clear.

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  • $\begingroup$ Oh right. But for a, is it also the same as using the formula $n!/(n_1 ! \times n_2 ! \times n_3 ! \times ... \times n_{12}!)$ where $n_1=6, n_2=4$, and the others are $1!$ (since they're all distinct)? $\endgroup$
    – clueless
    Jan 19 '19 at 15:52
  • $\begingroup$ And for b, the answer should be $(15!)(6!)$. Is this correct? (assuming the first interpretation is intended) $\endgroup$
    – clueless
    Jan 19 '19 at 16:01
  • $\begingroup$ @clueless You are correct on both counts. $\endgroup$ Jan 19 '19 at 16:04
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    $\begingroup$ Thank you so much! $\endgroup$
    – clueless
    Jan 19 '19 at 16:07
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The number of different permutations of $n$ objects, where $n_{1}$ are of one kind, $n_{2}$ are a different kind $\dots$ and there are $k$ different kinds is: $$ \frac{n!}{n_{1}! \times n_{2}! \times \dots \times n_{k}!} $$

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