SOLUTION: 2 math books,4 science books and 3 literature books are to be arranged on a shelf. In how many ways can this be done if the 2 math books are not together?

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Question 1102309: 2 math books,4 science books and 3 literature books are to be arranged on a shelf. In how many ways can this be done if the 2 math books are not together?
Found 2 solutions by math_helper, ikleyn:
Answer by math_helper(2461)   (Show Source):
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Ignoring the math book restriction, there are 9! = 362880 unique ways of arranging the books.
Now we need to divide that by the number of arrangements where the two math books are together:
There are 8 ways for m1m2 to appear and 8 ways for m2m1 to appear, so we need to divide by 16:
362880/16 =
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EDIT 11/25: Yes, my answer over-deducts (i.e. dividing by 16 takes out too many arrangements) for the M1M2, M2M1 cases. 9! - 2*8! removes the arrangements properly for M1M2 & M2M1 together. [ If I had sanity-checked my answer I'd have seen that it is way too small. ]

Answer by ikleyn(52776)   (Show Source):
You can put this solution on YOUR website!
.
The condition omitted one important detail: the books are considered as DISTINCT and DISTINGUISHABLE.

So, I will assume that this EXTENDED condition is in the place.

      Then the answer is 9! - 2*8! = 282240.

Solution

The number under the question is equal to the number of ALL PERMUTATIONS of 9 = 2 + 4 + 3 books/objects MINUS the DOUBLED number of all permutations of 8 objects, where we consider the two MATH book standing together as ONE emerged object with two states M1M2 and M2M1.