The task is to distribute 10 firms in 11 avalible spaces. The solution says the following: 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. The result is 10×10×9×8…×1=10×10!=36,288,000. I think the result should be different, since first firm has 11 avalible options (and not 10 as mentioned in solution). So the final result should be 11! = 39,916,800, or may be I am wrong. Could you please clarify this solution? Thank you in advance Andro

I think this problem can be interpreted in different ways. First interpretation: DB's place was already assigned (so there is one specific room that remains empty) and one of the other companies can use 2 rooms instead of 1: in this case we can arrange the other 10 companies in the 10 rooms left (10! permutations), then assign to one of the 10 companies the extra room. So the result is 10 x 10!. Second interpretation: DB's place wasn't assigned yet and one of the other companies can use 2 rooms instead of 1. In this case, we can arrange 10 companies in 11 rooms (variations without repetition with n=11 and p=10, so 11!), then assign the extra room to one of the 10 companies. So the result is 10 x 11!; alternative way to see it: we fix the extra room at a specific number from 1 to 11, then we have 10! permutations to arrange the 10 companies in the 10 places left, then we have to assign the extra room to one of the 10 companies. So we have 11 x 10! x 10 possibilities, that in fact is equal to 10 x 11!. Third interpretation: DB's place was already assigned (so there is one specific room that remains empty) and every company can use just one room (excluding DB's place). In this case, we just have the permutations of 10 companies in 10 places. So the result is 10!. Fourth interpretation: DB's place wasn't assigned yet and every company can use just one room. In this case, we can arrange 10 companies in 11 rooms (variations without repetition with n=11 and p=10). So the result is 11!; alternative way to see it: we fix the empty room at a specific number from 1 to 11, then we have 10! permutations to arrange 10 companies in 10 places. So the result is 11 x 10!, that in fact is equal to 11!. Fifth interpratation: DB's place was already assigned (so there is one specific room that remains empty) and every company can use just one room (including DB's place). This is equal to the fourth interpretation! I think that this problem's interpretation is the first one, since the result is 10 x 10!.