Tell me how to figure out these probabilities

In how many different orders can 9 people stand in line?

In how many ways can 4 people be seated in a row of 12 chairs?

Thank you

For the first problem, we have nine people and want to arrange them in a line. For the first position there are 9 choices, for the second position 8, the third 7, etc. on down to 2 and 1. Thus there are
9*8*7*...*2*1 ways to arrange 9 people in a line.

For the second problem, the first person has 12 choices to be seated, the second has 11, the third 10 and the fourth 9. Another way to think of the problem is to ask how many ways could we select 8 empty chairs from the 12? Of the 4 chairs that were not selected, there are 4! ways to seat the people, thus the number would be
4! times 12 choose 8 = 4!*(12!/(4!8!)). Verify that both methods are equivalent.

9 people can stand in line 81 ways-I think-I'm not sure though
And i don't know the answer to the second one. My answer is alot

Emily, I know that I am responding very late, but I do not think that 81 is the answer...I know that it is way more than 100

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  1. six people seated at a table. If two tables are put together,10people can be seated.How many tables are needed to make a long table that will seat 22 people?

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