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The class consists of 7 girls and 11 boys. How many ways are there to seat them behind 12 double desks? Every girl should sit next to the boy. Boys can sit together or one by one.

My logic is multiply 3 big numbers:

  • we can put the girls at the desks: $24 * 22 * 20 * 18 * 16 * 14 * 12$

  • we should put 6 free places on desks where we have no gils: $10 * 9 * 8 * 7 * 6 * 5$

  • after that we have $11!$ combinations for boys

Is it correct?

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  • $\begingroup$ The order in which we select the free places does not matter. $\endgroup$ – N. F. Taussig Mar 25 '19 at 19:27
  • $\begingroup$ @N.F.Taussig, what do you mean? We have 5 desks without girls. It's 10 places, is not? $\endgroup$ – Evgeny Mar 25 '19 at 19:30
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I believe it should be solved in the following way:

Girls have 12 desks to choose (left or right): $$ ^{12}P_{7} \times 2^{7} $$

No girls alone mean that 7 boys are with them in the 7 desks that already have girls: $$ ^{11}P_{7} $$

The rest of the boys (4) are in the remaing 10 places: $$ ^{10}P_{4} $$

Putting everything together gives: $$ ^{12}P_{7} \times 2^{7} \times ^{11}P_{7} \times ^{10}P_{4} $$

I assumed that the desks are distinguishable, e.g., being in the front left desk is different from being in the middle right desk.

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No, selecting the free places should not depend on the order, so it is ${10 \choose 6}$. You have ten chairs that could be free and six of them will be.

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