Combinations with an empty seat

Question

I'm trying to solve a question regarding permutations and combinations.

This year, you are helping organize your college’s career fest. There are 11 companies which are participating, and you have just enough room fit all of them. How many ways can you arrange the various firms, assuming…:

1) … Deutsche Bank representatives cancel, so you can give the additional space to one of the other companies?

My approach to the question was that we will first calculate the number of ways 10 seats will be selected from the given 11 chairs which will be 11C10 i.e. 11

We will then find out the total permutations which will be 10! therefore the answer will be 11 x 10!.

In the book, the answer is given as 10 x 10! with the explanation as:

We have 10 firms, which need to fill out 11 spots. Then, if we start filling up the room in some specific order, then there are going to be 10 options for who gets the first position. Since any firm can be given the additional space provided by DB’s withdrawal, then there are once again 10 options for the second spot. Then, there would be 9 different options for the third and so on. This results in having 10 × 10 × 9 × 8 … × 1 = 10 × 10! = 36,288,000 many options to arrange the firms.

I think the answer is wrong but would appreciate a second opinion

Answer 1

Another way to think about it is to assign DB anyway, yielding $11!$ assignments. Now reassign DB's slot to one of the other 10 companies, in 10 ways. But now divide the count by 2 because you have counted every arrangement twice. The final answer would then be $11!10/2=5\cdot 11!$, which differs from both yours and the book's.

Equivalently, choose which of the 10 companies will get 2 slots, choose 2 slots from 11 for that company, and assign the other 9 companies to the other 9 slots: $$10\binom{11}{2}9!=10\cdot\frac{11\cdot 10}{2}\cdot9!=5\cdot 11!$$

Answer 2

Your method seems not to take into account the "you can give the additional space to one of the other companies" bit. If the 11th space were simply left empty, your method would be correct. And in fact the answer would simplify to 11! which makes sense as you can simply treat the empty space as another company.

The correct answer depends on how we interpret "you can give the additional space to one of the other companies":

(i) The two spaces do not have to be next to each other

In this case the answer is $5 \times 11!$ as stated by Rob Pratt above.

(ii) The two spaces must be next to each other

In this case we can first decide which company gets the two spaces, giving 10 possibilities. We can then arrange the companies in 10! ways, giving an answer of $10 \times 10!$

This is the same answer as in the book. However, it seems to me that the book's reasoning is wrong: I fail to see why the number of options for the third space would reduce from 10 to 9 if the first two spaces were filled by different companies. Also the book's method doesn't seem to take into account any supposition that the two spaces are next to each other, and such a supposition is I think necessary for the book's answer to be correct.