0

Question: How many ways can you arrange a group of $10$ students given that $3$ of them cannot sit next to each other (they cannot sit as a group of $3$ nor can $2$ of them sit next to each other $-$ the three students need to be separated completely).

My attempt: Naturally, I went to counting possible cases and I decided to visualise different combinations if they were arranged in a line. I ended up with $15$ combinations that they could sit in a line such that they are unique and follow the condition if each line was a circle. From there it was pretty easy to see that $\dfrac{15 \times 3! \times 7!}{10}$ was the resulting answer. I have tried to justify the $15$ without counting situations, although I have been unsuccessful. Is this answer correct? If it is, how can I properly justify it? If it isn't, how would I approach a problem like this?

Any guidances or help would be greatly appreciated!

EDIT: I know for a fact $15$ is incorrect. I feel like it may be $12$ but still I'm not entirely sure.

LogicAndTruth
  • 609
  • 6
    Are they in a line or in a circle? If a line, then arrange the three troublesome students as $_X_Y_Z_$ and now populate those gaps. Note that the gap before $X$ and the gap after $Z$ might be empty but the internal ones can't be. If it's a circle, it's the same only now the gap before $X$ is the same as the gap after $Z$ and it can not be empty. – lulu Aug 12 '20 at 11:31
  • Maybe you look at this problem: https://math.stackexchange.com/questions/3774236/n-circular-arrangement-problem/3774369#3774369 – Christian Blatter Aug 12 '20 at 12:29
  • Stop treating this math problem as a guessing game. Don't put forth a number unless you have justification. – Jake Mirra Aug 12 '20 at 12:41

4 Answers4

2

(1) Assuming they in a line, not a circle, then imagine $7$ stars a row (the "other" students). Among the $8$ gaps in between or outside of those stars, choose $3$ for the locations of the special $3$ students. This can be done in $C(8,3)=56$ ways.

(2) Now order (say, left to right) both of the two subsets of students arbitrarily, this can be done $(7!)(3!)$ ways.

Each admissible arrangement of students corresponds to a unique sequence of choices in (1) and (2). So there are $56(7!)(3!)$ ways to do it.

Ned
  • 4,012
0

If the 10 students sit in a circle thats my solution:

First lets see in how many cases $3$ of our special students sit next to each. Thats $10$.

Now we want to know in how many cases $2$ sit next to each other. If we had picked our $2$ neighboured seats then there are $$ \frac{3!}{(3-2)!} = 6 $$ to sit our 3 students onto them.

Now to place this "double seat" we have $10$ possibilities and after we placed that one since we dont want a triple seat (we covered that case already) we have $6$ possibilities left for the third student to sit. That makes a total of $6^2 \cdot 10 = 360$ possibilities for this case.

So we got a total of $370$ 'bad' cases.

Now to sit the $3$ students down without restriction we have

$$ \frac{10!}{(10-3)!} $$

cases. lets substract the bad ones so we get $\frac{10!}{7!} - 370$ 'good' cases. After having sat the 3 students down according to the restriction, we can now distribute the remaining $7$ students with a total of $7!$ possibilities. So we get

$$ 10! - 370\cdot 7! $$

crush3dice
  • 317
0

Chancing upon this question, here is an answer using successive placement

First arrange the "ordinary" people, and use successive placement to place the three special people one by one in the (reducing) gaps, thus

ans = $7!\cdot8\cdot7\cdot6$ ways

true blue anil
  • 48,250
-1

The assumption in crush3dice's answer that the students sit in a circle is just as reasonable as the assumption that I am going to make, that the students sit in a row of chairs numbered, from left to right as 1, 2, ..., 10.

My approach is to favor brute force over elegance of thought in this one-off (small) setting. If instead the # of chairs was (for example) 1000, then I would look for a pattern and then try for elegance.

Suppose that you select chairs 1, 3, 5, and decide that the 3 troublesome students (refer to them as a, b, and c) will sit in those 3 chairs. There are 3! ways of permuting these 3 students among these 3 chairs and 7! ways of permuting the other 7 students among the other 7 chairs.

Therefore, the exact answer is $3! \times 7! \times T$, where $T$ represents the total # of ways of selecting 3 chairs out of 10 so that none of the three chairs is next to each other.

Now for the brute force:

Rather than try for elegant analysis, why not manually enumerate $T$.

Let $L$ = the # of the lowest of the three chairs.
Let $M$ = the # of the middle of the three chairs.
Let $N$ = the # of the largest of the three chairs.

Clearly, $1 \leq L \leq 6.$
That is if $L = 7$, there is no way of selecting 2 chairs among chairs 8, 9, and 10 so that the 2 chairs are apart both from each other and chair 7.

If $L$ and $M$ are set then $N$ must be in $\{(M+2), \cdots, 10\}.$

If $L$ is set, then $M$ must (similarly) be in $\{(L+2), \cdots, 8\}.$

Since you know that $L$ is in $\{1,2,3,4,5,6\},$ then all you have to do to enumerate $T$ is to count.

Addendum : elegance rears its ugly head

Since the following idea just came to me, I might as well share it. There is a shortcut to enumerating $T.$

As analyzed, $M$ must range from 3 through 8, inclusive.

For any specific $M$ in this range, $L$ must range from 1 through $M-2$ and $N$ must range from $M+2$ through 10.

Therefore, $T = \sum_{M=3}^8 \,[(M-2) \times (9-M)].$

Addendum-2 : A symmetry argument.

By symmetry, the # of arrangements with $M=3$ equals the # of arrangements with $M=8.$ A similar argument relates $M=4$ to $M=7$ and $M=5$ to $M=6.$

Therefore, $T = 2 \times \left\{\sum_{M=3}^5 \,[(M-2) \times (9-M)]\right\}.$

user2661923
  • 42,303
  • 3
  • 21
  • 46