1. ## The Counting Principle

My AP Statistics teacher gave my class two sheets of problems today(first day of class) to see how we could do. I haven't done problems like these for a year or two so I want to make sure I'm on track.

I want to make sure these three in particular are correct:

1.How many ways can 8 members of a family be seated side-by-side in a movie theater if the father is seated on the aisle?
My attempt:
$1(7!)=5040$
I represented the father as $1$, I know $7!$ and $7!(1)$ are the same. Is my answer correct?

2. How many license plate numbers consisting of three letters followed by three numbers are possible when repetition is allowed?
My attempt:
$(26^3)(10^3)=17,576,000$

3. How many combinations are possible using the information in problem 7(the above problem) if no repetition is allowed?
My attempt:
$(26)(25)((24)(10)(9)(8)=11,232,000$

I vaguely remember a was to express a factorial that doesn't multiply all the way to 0. Like this: $(26)(25)(24)$ doesn't equal $26!$, so how do I express it?

I know these aren't the most challenging problems but I don't want to make any silly mistakes. Thank you!

2. Originally Posted by agm
My AP Statistics teacher gave my class two sheets of problems today(first day of class) to see how we could do. I haven't done problems like these for a year or two so I want to make sure I'm on track.

I want to make sure these three in particular are correct:

1.How many ways can 8 members of a family be seated side-by-side in a movie theater if the father is seated on the aisle?
My attempt:
$1(7!)=5040$
I represented the father as $1$, I know $7!$ and $7!(1)$ are the same. Is my answer correct?

2. How many license plate numbers consisting of three letters followed by three numbers are possible when repetition is allowed?
My attempt:
$(26^3)(10^3)=17,576,000$

3. How many combinations are possible using the information in problem 7(the above problem) if no repetition is allowed?
My attempt:
$(26)(25)((24)(10)(9)(8)=11,232,000$
These are correct in concept( I did not check the caculations. I never want students to give me numbers worked out).

I will point out the in #3 you have two permutations:
$P(26,3)\cdot P(10,3)=\frac{26!}{(26-3)!}\frac{10!}{(10-3)!}$

3. Originally Posted by Plato
These are correct in concept( I did not check the caculations. I never want students to give me numbers worked out).

I will point out the in #3 you have two permutations:
$P(26,3)\cdot P(10,3)=\frac{26!}{(26-3)!}\frac{10!}{(10-3)!}$
Oh yes, I remember that now. Thank you for the help!