A (small) theatre contains 10 seats. The seats are numbered from 1 to 10,

Question

A (small) theatre contains 10 seats. The seats are numbered from 1 to 10, but not in order. Alice, Bob and Carol have tickets for seats 1, 2 and 3 respectively, and Damien, Eve, Francis, and Gertrude each have tickets without an assigned seat. Unfortunately they all arrive late and the lights are dim, so they can’t see what seat they are sitting in.

a) In how many ways can they be seated?

b) In how many ways can they be seated such that Alice is in her correct seat (eg, seat number 1)?

c) In how many ways can they be seated such that Alice, Bob and Carol are all in their correct seats?

d) In how many ways can they be seated such that none of Alice, Bob, Carol are in their correct seats?

What I did for part A is The first person has 10 seats to choose from, then the second person has 9 and so on. I got $10\times9\times8\times7\times6\times5\times4$ as the answer.

For part B I said alice has 1 seat to choose from and then did the rest like part A so I got $1\times9\times8\times7\times6\times5\times4$.

For C I did the same as B but with 3 people and got $1\times1\times1\times7\times6\times5\times4$ but this doesn't seem right to me.

A bit confused on how to do the rest.

Answer 1

"For c) I did the same as b) but with 3 people and got $7^{\underline{4}}=7\cdot6\cdot5\cdot4$ but this doesn't seem right to me"

Your first three parts are in fact all correct (though they suffer from a lack of convenient notation. If there were 50 people and 80 seats you don't want to waste time writing out all of the numbers in the product. Recommend using one of the following falling factorial notations $(7)_4, 7^{\underline{4}}, \frac{7!}{3!}, ~_7P_4, P(7,4),\dots$ but whichever you use make sure your audience understands the notation)

(Note on falling factorial notation: in all of the above, $(n)_r=n^{\underline{r}}=P(n,r)=\dots=\underbrace{n(n-1)(n-2)\cdots(n-r+2)(n-r+1)}_{r~\text{terms in product starting with}~n}$)

"A bit confused on how to do the rest"

The first three parts are most of the calculations needed to complete the fourth part. Recognize that by symmetry the answer to part b) is also the answer to the question of how many ways Bob is seated in his correct seat as well as the answer to how many ways Carol is in her correct seat.

Use inclusion-exclusion to figure out answer to "In how many ways can they be seated such that at least one of them is in their correct seat." Letting $A,B,C$ represent the events that Alice, Bob, and Carol made it to their correct seats respectively, and letting $\Omega$ represent the sample space i.e. the number of ways in which we don't care who did or didn't make it to their seat, we are tasked with finding the number of ways that none of them made it to their seat. I.e. $|A^c\cap B^c\cap C^c|$

Note that $|A^c\cap B^c\cap C^c| = |\Omega\setminus(A\cup B\cup C)|=|\Omega|-|A\cup B\cup C|$

$=|\Omega|-|A|-|B|-|C|+|A\cap B|+|A\cap C|+|B\cap C|-|A\cap B\cap C|$

You've already found $|\Omega|,|A|,|A\cap B\cap C|$ and by symmetry found $|B|$ and $|C|$. The only information you are missing is $|A\cap B|,|A\cap C|,|B\cap C|$. A similar symmetry argument will reduce the effort it takes to find those.