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$11$ papers are set for an examination in which two are of mathematics. Number of ways in which papers can be arranged so that mathematics papers do not come together?

There are $11!$ ways in total. If we treat two mathematics papers as one unit then there will be $10!$ ways in which they will be always together.

Therefore, the answer should be $11!-10!$, but the answer is wrong. Please help.

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  • $\begingroup$ Total possible ways to arrange 11 papers is $\frac{11!}{2}$ as the order of 2 mathematics paper is important. $\endgroup$ – Murtuza Vadharia May 1 '16 at 15:29
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HINT:

The order of the $2$ mathematics papers is significant.

Multiply $10!$ by the number of ways in which you can arrange those items.

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  • $\begingroup$ the answer is 9!*10! $\endgroup$ – kadir May 1 '16 at 15:04
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    $\begingroup$ @kadir: So have you tried to combine my hint into your answer and see if you get the same? $\endgroup$ – barak manos May 1 '16 at 15:04
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    $\begingroup$ @kadir Notice that $11! - 2 \cdot 10! = 11 \cdot 10! - 2 \cdot 10! = (11 - 2) \cdot 10! = 9 \cdot 10!$. The answer is clearly not $9! \cdot 10! > 11!$. $\endgroup$ – N. F. Taussig May 1 '16 at 15:06
  • $\begingroup$ there are 10 positions in which we can fit two math papers and this can be done in 90 ways...i think $\endgroup$ – kadir May 1 '16 at 15:11
  • $\begingroup$ @kadir: Quoting from your question: "If we treat two mathematics papers as one unit then there will be $10!$". Good, but how many different ways are there to treat those two paper as one unit? (HINT: more than one). $\endgroup$ – barak manos May 1 '16 at 15:17

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