# [SOLVED] fundamental counting principal.

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• May 17th 2009, 10:52 AM
brentwoodbc
[SOLVED] fundamental counting principal.
I am alright at the questions where you are told how many objects there are, and are asked how many combinations there are of those objects. My question is what about when it says something like....

"A car liscence plate consists of up to 6 characters" with any combo of the letters a-1 and numbers 0-9"

or

A room key for a hotel has a 5 by 10 array where a hole is either punched or let solid, how many different room keys are possible."

*I'm not necessarily asking you to solve both questions, I just dont know how you do a question where it could be 1 digit on a plate or 6.

Thanks.
• May 17th 2009, 10:55 AM
Jen
Think of all the ways that you can have a license plate with one letter/number on it, how many ways for two letter/numbers, how many ways for 3...

and so on. Then add them together.

Hope that helps
• May 17th 2009, 11:00 AM
brentwoodbc
This is in the fundamental counting part of the text book but I thought it might be done using factorials. I tried (26+10)!/(35!) + (36)!/(34)!
but that was taking for ever and gave me the wrong answer.(Crying)
• May 17th 2009, 11:17 AM
Jen
$\displaystyle 36 \choose 1$, will give you the number of ways that you can choose one letter or number out of 36 to put on a license plate.

Do this for the license plate with 2 elements and 3 and so on and sum them.

I think this should give you a total number of ways that you can make a license plate. (Without repeat numbers or letters)

(I am not so great at combinatorics though)

Hope this helps.
• May 17th 2009, 11:17 AM
Plato
Quote:

Originally Posted by brentwoodbc
This is in the fundamental counting part of the text book but I thought it might be done using factorials. I tried (26+10)!/(35!) + (36)!/(34)! but that was taking for ever and gave me the wrong answer.

I find the way your textbook words questions makes it very hard to understand.
However, I think that the answer should be $\displaystyle \sum\limits_{k = 1}^6 {36^k }$.

You see the number of plates with a combination of four letters or digits is $\displaystyle 36^4$.

So we sum from one to six.
• May 17th 2009, 11:21 AM
Jen
If you can have repeats on a license plate then the first slot has 36 options, the second slot has 36 option and so on

so

lic. with one letter/number 36

Lic. with two letter/number 36*36

Lic with 3 letter/number 36*36*36

and so on, then add them up.
• May 17th 2009, 11:22 AM
Jen
Haha, which is just what Plato said.

Only mine doesn't look as nice. :)
• May 17th 2009, 11:27 AM
brentwoodbc
I'll rewrite straight from the text book.

"assume a car license plate consists of up to 6 characters. Each character can be any of the letters a-z, or any digit 0-9 how many license plates are possible?

plates is 223897*6116

In a Hong kong hotel a room key is a card. The card has positions for holes that form a 5 by 10 array. Each position in the array is punched or left blank. How many different keys are possible?

and for the keys is 1.126*10^15
• May 17th 2009, 11:30 AM
Jen
Ahhh, so the text doesn't use the word "combo".
• May 17th 2009, 11:32 AM
brentwoodbc
no(Worried)
• May 17th 2009, 11:34 AM
Jen
Then the solution that I discribed and Plato so nicely state in summation notation should work.

The word "combination" has a very specific meaning in counting principles.
• May 17th 2009, 11:37 AM
brentwoodbc
sorry I havent done "combinations" yet I thought it was the same thing.

I did 36+36^2.....and got 223897116 :/
• May 17th 2009, 11:39 AM
HallsofIvy
Quote:

Originally Posted by brentwoodbc
I'll rewrite straight from the text book.

"assume a car license plate consists of up to 6 characters. Each character can be any of the letters a-z, or any digit 0-9 how many license plates are possible?

Then characters [b]can[b] be repeated. There are 26 letters and 10 digits for a total of 36 possible characters.
There are 36 possible 1 digit character, $\displaystyle 36(36)= 36^2$ 2 character plates, [tex]36^3[tex] 3 character plates, etc. The number of plates with up to 6 plates is $\displaystyle 36+ 36^2+ 36^3+ 36^4+ 36^5+ 36^6$. That could be calculated directly or done using the formula for a finite geometric series. I get 2238976116.

Quote:

plates is 223897*6116

In a Hong kong hotel a room key is a card. The card has positions for holes that form a 5 by 10 array. Each position in the array is punched or left blank. How many different keys are possible?

and for the keys is 1.126*10^15
Each hole has two "positions" and there are 50 holes: $\displaystyle 2^{50}$= 1125899906842624 which is what you give to 4 significant figures.
• May 17th 2009, 11:41 AM
brentwoodbc
since the text gave an answer of 2 things multiplied by each another I tried 26+26^2...times 10+10^2.... and that doesnt work either.(Headbang)
• May 17th 2009, 11:43 AM
brentwoodbc
thanks everyone, I really appreciate the help. maybe it's an error in the text, it wouldnt be the first time.
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