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I study maths as a hobby. I have come across this problem:

On a shelf there are 4 saucers of different colours and 4 matching cups. In how many ways can the cups be arranged on the saucer so that no cup is on a matching saucer?

I start off by saying the first cup can be placed on any of 3 saucers. For the second cup, there are 3 choices, unless the first cup was placed on the second cup’s matching saucer, in which case there are only 2 choices. That gives 8 outcomes so far. But the answer in the book is 9. So I know my method is wrong.

I have seen similar problems posted on here but the solutions were too complex for me.

Steblo
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2 Answers2

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There are $3$ cups on which you can place on saucer $4$. For concreteness, let's say you put cup $3$ there. Now, you might place cup $4$ on saucer $3$, and then only cups $1$ and $2$ and saucers $1$ and $2$ remain, and there's only one way to place the cups on non-matching saucers, so in this case we have $3$ possibilities.

Alternatively, cup $4$ might be placed on either saucer $1$ or saucer $2$. Now we still have to place cups $1$ and $2$ and saucers $2$ and $3$ are available. Again, there is only one way to place the cups on non-matching saucers, so we have $2\cdot3=6$ possibilities in this case, and $9$ overall.

The generalization of this line of reasoning to $n$ cups and saucers is given here.

saulspatz
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Place cup A on saucer B say, there are $3$ ways to do this. Place cup B on any remaining saucer, there are again $3$ ways to do this. The last $2$ cups are forced onto specific saucers, hence $9$ ways in total.

JMP
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