• December 24th 2012, 11:51 AM
lhurlbert
I was posed the question
"how many license plates contain at most 3 numerals followed by exactly 4 distinct letters"

I figured

3x26!-22! +
2x26!-22! +
1x26!-22!

How is this expressable or ssimplified and am I even correct
• December 24th 2012, 12:01 PM
Plato
Quote:

Originally Posted by lhurlbert
I was posed the question
"how many license plates contain at most 3 numerals followed by exactly 4 distinct letters"

I figured

3x26!-22! +
2x26!-22! +
1x26!-22!

That does not work.

There are $_{26}P_4=(26)(25)(24)(23)$ ways to have exactly 4 distinct letters.

There are $1+10+100+1000$ ways to at most three numerals.
Don't forget that are ten numerals, and "at most three" means 0, 1, 2, or 3.
• December 24th 2012, 09:22 PM
Soroban
Hello, lhurlbert!

Quote:

How many license plates contain at most 3 numerals followed by exactly 4 distinct letters?

It starts with at most 3 digits.

It can have no digits: 1 way.
It can have any number from 1 to 999.
. . Hence, there are 1000 choices for the numerals.

$\text{There are: }\,26\cdot25\cdot24\cdot23 \:=\:358,\!800\text{ possible four-letter "words"}$
. . $\text{with distinct letters.}$

$\text{Therefore, there are: }\,(1,\!000)(358,\!800) \:=\:358,\!800,\!000\text{ possible license plates.}$

• December 25th 2012, 04:59 AM
Plato
Quote:

Originally Posted by lhurlbert
I was posed the question
"how many license plates contain at most 3 numerals followed by exactly 4 distinct letters"

Quote:

Originally Posted by Soroban
It starts with at most 3 digits.
It can have no digits: 1 way.
It can have any number from 1 to 999.
. . Hence, there are 1000 choices for the numerals.

As you can see the answer above differs from the one I gave.
Here is why: I read the question in such a way that the plates $90ABXT~\&~090ABXT$ are different.
In other words: There is 1 way for no digits.
There are 10 ways for one digit.
There are 100 ways for two digits.
There are 1000 ways for three digits.