4
$\begingroup$

We have 4 French, 2 Spanish and 3 German books. The French and German books have to be together. All the books are unique, they came from the same country but have different volume numbers.

I combined the 3 German and 4 French and got $7!$, and then I added the 2 Spanish books. Since they are unique, lets assume C = the combined books. I have SSC x2, SCS x2, CSS x2, so I have 6 ways of arranging them. I then did $7! \cdot 6$

Is this the right way of doing the problem or am I missing a few things?

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    What have you considered so far? – 2017-02-16
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    Sorry if my question was lacking in my own answers, I'm a bit new to questioning. Edited it to included what I have considered. – 2017-02-16
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    Is the answer 24? – 2017-02-16
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    That would be too small. My answer was 30,240 ways because of 7! x 6, but I am not entirely sure on that answer – 2017-02-16
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    @MrAP, we do have 9 objects here. If it were 9!, then the equation would be 362,880 ways. That's a big number. But 30,240 is significantly smaller than 362,880 – 2017-02-16
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    Hey, no problem! I see some people getting their questions downvoted or closed, and was trying to help you prevent it. – 2017-02-16
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    I get the answer to be 2.4.3.3.2.2.1.1=288 – 2017-02-16
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    Can somebody explain this question to me? – 2017-02-16

1 Answers 1

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Yes, your working looks good. The $6$ can be regarded as $3!$, the ways of arranging three objects (one of which is a composite object).

An aside: $7!3!=30240$ is the smallest 4-perfect number.