In how many ways can $14$ sixth graders and $10$ fifth graders be arranged in a line so that no two fifth graders occupy consecutive positions?

Question

My approach :

I imagined the $10$ fifth graders are standing in a line and in between every two of them there is one sixth grader standing. So the picture is basically this:

FSFSFSFSFSFSFSFSFSF

There are still $5$ sixth graders left. They have $11$ possible places to go (in between the fifth graders or at the ends). What I mean is the line may be like this:

SSSSSFSFSFSFSFSFSFSFSFSF

or may also be this:

SSSFSFSFSFSFSFSFSSSFSFSF

So I calculated there are $11^5$ such combinations because of that. Then the total number of lines that could be formed are: $11^5 \times 14! \times 10!$

And I know I am totally wrong. I just want to know why. the inspiration of solving this question in this method lies in the selected answer of another question I asked:

There are 3 candidates for professorship and one is to be elected by the votes of 5 voters. In how many ways the votes can be given?

Answer 1

Your mistake is that you are double-counting: suppose that you would just have two sixth graders left to place. You say that since the first has $11$ options to go, and the second has $11$ options to go as well, there are $11^2$ options. But, note that if the first goes in place $5$, and the second in place $7$, then you end up with the same configuration (in terms of initially indistinguishable $F$'s and $S$'s) as when the first goes in place $7$, and the second in place $5$.

One way to try and repair your approach, then, is to distinguish between the leftover sixth grader. However, when you do so, notice that whenever several sixth graders end up in the same of the original $11$ positions, you then need to order them one way or the other ... as such, as NewGuy points out, after placing the first of the leftover sixth graders, you really have $12$ spots left for the second, etc. Also, since you are distinguishing all sixth graders at the end, you need to 'indistinguish' the $5$ additional sixth graders after having placed them, meaning you get:

$$\frac{11\cdot 12\cdot13\cdot14\cdot15}{5!}={15\choose 5}$$

ways to get the number of arrangements of indistinguishable $F$'s and $S$'s, for a total of:

$${15\choose 5}\cdot 14! \cdot 10!$$

But, here is what I think is a somewhat simpler method:

First, line up the $14$ sixth graders, and leave room between and next to them for fifth graders to go:

$$*S*S*S*S*S*S*S*S*S*S*S*S*S*$$

OK, so you have $15$ places where the fifth graders can go. However, since no two fifth graders can be next to each other, that means that no two fifth graders can be in the place. This means that you need to pick $10$ different places out of $15$, i.e. There are $15 \choose {10}$ ways to get valid arrangements of otherwise indistinguishable fifth graders and sixth graders. And aince they are to be distinguishjed, the number of different arrangements is:

$${15 \choose 10} \cdot 14! \cdot 10!$$

Since ${15 \choose 5} = {15 \choose 10}$, this is the same answer as before.

Answer 2

To be found is the number of sums $$a_1+\dots+a_{11}=14$$ where $a_1,a_{11}$ are nonnegative integers and where the other $a_i$ are positive integers.

Here $a_1$ represents the number sixth-graders on the left of the utmost left fifth-grader, $a_2$ represents the number sixth-graders between the utmost left fifth-grader and the fifth-grader at the utmost left except for one, et cetera.

That comes to the same as finding the number of sums $$b_1+\dots+b_{11}=14-9=5$$ where the $b_i$ are nonnegative integers.

This can be solved with stars and bars.

Further there are $10!$ possible orders for the fifth-graders and there are $14!$ possible orders for the sixth-graders.