I have tried to solve it. But whatever way I came across, I ended up double counting. please help. thanks.


closed as off-topic by Xander Henderson, Morgan Rodgers, Shailesh, Saad, JonMark Perry Mar 3 '18 at 6:12

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  • $\begingroup$ Can you explain your approach, and how it lead to double counting? Also, could you please include the text of your question in the question? One should be able to know what is going on without reference to the title. $\endgroup$ – Xander Henderson Mar 3 '18 at 4:30
  • $\begingroup$ @Xander Henderson I was embarrassed to mention it. As those lacked logic. $\endgroup$ – abu obaida Mar 3 '18 at 5:35

I only sum up the already present solution in the comments:

  • 1st factor: $14!$ arrangements of $10$th graders
  • 2nd factor: $\binom{15}{10}$ possible choices of "slots" to put exactly one $9$th grader
  • 3rd factor: $10!$ arrangements of the $10$ $9$th graders within a given choice of slots

$$\mbox{Result: }14! \cdot \binom{15}{10} \cdot 10!$$


I actually came across a similar problem not too long ago.

The logic here is to first arrange the 14 tenth graders in a line. What we obtain is that there are 15 spaces between them (or at either end) where a ninth grader can go.

By looking at it this way, and since we don't want two ninth graders to be consecutive, we are in essence asking ourselves a new question: In how many ways can 10 ninth graders be placed into 15 different spots?

The answer to this will simply be:


  • 2
    $\begingroup$ 15 C 10 are the number of ways you can select 10 positions out of 15.it does not take into account the permutations of 10th and 9th grade students $\endgroup$ – NewGuy Mar 3 '18 at 5:25
  • $\begingroup$ 15C10. 14! is the answer, I think. could you complete ur answer. I think u forgot to permut it. $\endgroup$ – abu obaida Mar 3 '18 at 5:32
  • $\begingroup$ Perhaps I made an error somewhere. In the original question from which I borrowed the solution, the question assumed that the two different objects to be arranged were indistinguishable. $\endgroup$ – Bryden C Mar 3 '18 at 5:34
  • 2
    $\begingroup$ @abuobaida an answer of $\binom{15}{10}\cdot 14!$ would make sense as an answer if the tenth graders all had names and were different people but the ninth graders were all identical indistinguishable clones of one another. You'll need an additional factor of $10!$ if you want the ninthgraders to be distinct as well. $\endgroup$ – JMoravitz Mar 3 '18 at 5:43
  • $\begingroup$ @JMoravitz ur right. could u answer in detail. I will accept that. $\endgroup$ – abu obaida Mar 3 '18 at 5:59

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