In how many ways can 14 tenth graders and 10 ninth graders be arranged in a line so that no two ninth graders may occupy consecutive positions?

Question

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

Answer 1

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

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

Answer 2

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:

$${15}\choose{10}$$