Six papers are set in an examination of which two are mathematics. In how many ways can the examination papers be arranged if the mathematics papers are not to be together?
Suppose the 6 papers are A, B, C, D, M1, and M2, where M1 and M2 are the
two mathematics papers.
1. First we calculate the number of ways of arranging the 6 papers
regardless of whether M1 and M2 are together or not.
2. Then we calculate the number of those ways that the M1 and M2 are
together.
3. Then we subtract the result of 2 from the result of 1.
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1. There are 6 ways to choose the top paper. For each of those choices
there are 5 ways to choose the next paper and so on. So there are
6*5*4*3*2*1 or 6! ways to arrange them regardless of whether M1 and M2
come together or not.
2. When M1 and M2 come together, we can think of using a paper clip to
clip the two mathematics papers together. Then we have only 5 things,
A,B,C,D, and (M1M2) if we clip them together with M1 on top of M2,
or else we have A,B,C,D, and (M2M1) if we clip them together with M2
on top of M1. Each of those give 5! ways each. So that's 2*5! ways
the papers can be arranged with M1 and M2 paper-clipped together.
3. So the answer is 6! - 2*5! = 720 - 2*120 = 720 - 240 = 480
Edwin
