# The Counting Principle

• Aug 25th 2009, 02:29 PM
agm
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:
$\displaystyle 1(7!)=5040$
I represented the father as $\displaystyle 1$, I know $\displaystyle 7!$ and $\displaystyle 7!(1)$ 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:
$\displaystyle (26^3)(10^3)=17,576,000$

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

I vaguely remember a was to express a factorial that doesn't multiply all the way to 0. Like this: $\displaystyle (26)(25)(24)$ doesn't equal $\displaystyle 26!$, 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! :)
• Aug 25th 2009, 02:43 PM
Plato
Quote:

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:
$\displaystyle 1(7!)=5040$
I represented the father as $\displaystyle 1$, I know $\displaystyle 7!$ and $\displaystyle 7!(1)$ 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:
$\displaystyle (26^3)(10^3)=17,576,000$

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

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:
$\displaystyle P(26,3)\cdot P(10,3)=\frac{26!}{(26-3)!}\frac{10!}{(10-3)!}$
• Aug 25th 2009, 03:04 PM
agm
Quote:

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:
$\displaystyle P(26,3)\cdot P(10,3)=\frac{26!}{(26-3)!}\frac{10!}{(10-3)!}$

Oh yes, I remember that now. Thank you for the help!