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There are $8$ boys and $6$ girls. In how many ways can they line up so that the front and end of the line are occupied by boys?

Someone told me that the answer is $3832012800$. Am I missing something?

Would the solution not be:

We first choose the front and end of the line. Since they must both be occupied by boys, we have $8 \cdot 7$ possibilities. Then, there are $12$ people left, with no constraints for any of them. Thus, $8*7*12!$.

EDIT: I explained my solution, which I have been told is correct.

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    $\begingroup$ I agree with you. $\endgroup$ – Matthew Daly Aug 13 at 7:17
  • $\begingroup$ The number is $8\cdot 12!$. It's close, but the factor of 7 is certainly missing. $\endgroup$ – Matthew Daly Aug 13 at 7:23
  • $\begingroup$ Your answer is correct. However, you should explain your reasoning so that readers can discern any errors you may have made. This tutorial explains how to typeset mathematics on this site. Welcome to MathSE. $\endgroup$ – N. F. Taussig Aug 13 at 8:25
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This kind of question is rather silly, but to provide an answer to the question how it can be that somebody said the answer was $3832012800=8\times12!$ while the real answer is $8*7*12!$ is that they made a mistake. This kind of things happens. All the time.

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