2
$\begingroup$

DISCLAIMER: I might not be using the correct terms for what I'm trying to explain, sorry about that.

The problem is, there are $28$ chairs and $19$ people, how many permutations are possible? (The order of the people is important) Now I reckon to get the amount of the combinations of the $9$ empty chairs is $\binom{28}{9} = \frac{28!}{(28-9)!\cdot 9!}$ - am I correct so far? For every one of those combinations, there are $19!$ ways to sort the people on the chairs, right?

So the solution to the problem would be $\binom{28}{9}\cdot 19!$. If that isn't correct so far please explain how I would go about that.

To my actual question, is there a different way (apart from writing every possible permutation, hah) to calculate this, preferably without having to know how to calculate possible groups ignoring order (as I did for the empty chairs combinations)?

1 Answers 1

2

Your reasoning is solid. Here is another way of doing it:

You want to order $19$ students and $9$ indistinguishable chairs. If we pretend, for a second, that the chairs are numbered or something (so we actually care about their order, not just where between the students they appear, but also in relation to one another), then we just have $28$ distinguishable objects. Those can be ordered in $28!$ ways.

But the chairs are indistinguishable. That means that every combination that we care about is counted $9!$ times in the above argument (once for each order the chairs can be put in, within the "chair-spots" of the given ordering). Therefore the final answer is $\frac{28!}{9!}$. This quantity is also known as $_{28}P_{19}$, the same way $\binom{28}{9}$ is known as $_{28}C_9$.

  • 0
    Oh so the formula $\frac{n!}{(n-k)!}$ holds true for $n < k$? – 2017-01-30
  • 0
    @Skilly What are you talking about, $n$28$ things, and we want to pick out $19$ of them, caring about their order, but not the internal order of the rest. That means $n = 28$ and $k = 19$. – 2017-01-30
  • 0
    Ah yes sorry, I think I am slowly beginning to understand it. I've always assumed $k$ was the amount of different _places_ _things_ can go and $n$ how many different _things_ there are. I didn't think of interchanging the two, giving each student a chair instead of each chair a student in my case. I've wrapped my head around it now, thank you for answering my question! – 2017-01-30