How many combinations

[Nalani]

My son asked me a question that I'm not sure I gave him the right answer to.

How many different Licence Plates can be made with the format of 123-ABC?

This was my take on it.

Excel Workbook
	A	B	C	D
2	123-ABC			Total
3	1,000	26,000	676,000	17,576,000
Sheet1

Am I right or wrong.

 

[Nalani]

I found my answer here after searching Google.

Although they use only 2 alpha and 3 numeric, I believe my calculation is correct.

 

[shg]

I believe you are spot on.

 

[Nalani]

Thanks shg

It's been awhile since Math Class, thank goodness for Google.

 

[T. Valko]

My son asked me a question that I'm not sure I gave him the right answer to.

How many different Licence Plates can be made with the format of 123-ABC?

This was my take on it.
Sheet1

<TABLE style="PADDING-RIGHT: 2pt; PADDING-LEFT: 2pt; FONT-SIZE: 11pt; FONT-FAMILY: Calibri,Arial; BACKGROUND-COLOR: #ffffff" cellSpacing=0 cellPadding=0 border=1><COLGROUP><COL style="FONT-WEIGHT: bold; WIDTH: 30px"><COL style="WIDTH: 64px"><COL style="WIDTH: 64px"><COL style="WIDTH: 64px"><COL style="WIDTH: 71px"></COLGROUP><TBODY><TR style="FONT-WEIGHT: bold; FONT-SIZE: 8pt; BACKGROUND-COLOR: #cacaca; TEXT-ALIGN: center"><TD></TD><TD>A</TD><TD>B</TD><TD>C</TD><TD>D</TD></TR><TR style="HEIGHT: 22px"><TD style="FONT-SIZE: 8pt; BACKGROUND-COLOR: #cacaca; TEXT-ALIGN: center">2</TD><TD style="FONT-WEIGHT: bold">123-ABC</TD><TD></TD><TD></TD><TD style="FONT-WEIGHT: bold">Total</TD></TR><TR style="HEIGHT: 22px"><TD style="FONT-SIZE: 8pt; BACKGROUND-COLOR: #cacaca; TEXT-ALIGN: center">3</TD><TD style="TEXT-ALIGN: right">1,000</TD><TD style="TEXT-ALIGN: right">26,000</TD><TD style="TEXT-ALIGN: right">676,000</TD><TD style="TEXT-ALIGN: right">17,576,000</TD></TR></TBODY></TABLE>
<TABLE style="FONT-SIZE: 10pt; BORDER-LEFT-COLOR: #00ff00; BORDER-BOTTOM-COLOR: #00ff00; COLOR: #000000; BORDER-TOP-STYLE: groove; BORDER-TOP-COLOR: #00ff00; FONT-FAMILY: Arial; BORDER-RIGHT-STYLE: groove; BORDER-LEFT-STYLE: groove; BACKGROUND-COLOR: #fffcf9; BORDER-RIGHT-COLOR: #00ff00; BORDER-BOTTOM-STYLE: groove"><TBODY><TR><TD>Spreadsheet Formulas</TD></TR><TR><TD><TABLE style="FONT-SIZE: 9pt; FONT-FAMILY: Arial" cellSpacing=0 cellPadding=2 border=1><TBODY><TR style="FONT-SIZE: 10pt; BACKGROUND-COLOR: #cacaca"><TD>Cell</TD><TD>Formula</TD></TR><TR><TD>A3</TD><TD>=10*10*10</TD></TR><TR><TD>B3</TD><TD>=A3*26</TD></TR><TR><TD>C3</TD><TD>=B3*26</TD></TR><TR><TD>D3</TD><TD>=C3*26</TD></TR></TBODY></TABLE></TD></TR></TBODY></TABLE>

Excel tables to the web >> http://www.excel-jeanie-html.de/index.php?f=1" target="_blank"> Excel Jeanie HTML 4


Am I right or wrong.

26 letters choosing 3 =PERMUT(26,3) = 15,600
10 digits choosing 3 =PERMUT(10,3) = 720

15600*720 = 11,232,000

 

[Nalani]

Okay Biff, now you got me really confused.

With your formula the total is 6.344 millon short of what I had and what I think the link that I posted would have come out to be if they used 3 alpha's.

Could you explain, please.

I am assuming that there is no restrictions on the sequence.

 

[T. Valko]

Okay Biff, now you got me really confused.

With your formula the total is 6.344 millon short of what I had and what I think the link that I posted would have come out to be if they used 3 alpha's.

Could you explain, please.

I am assuming that there is no restrictions on the sequence.

First off, I posted this

for a reason!

Letters A to Z = 26 possible choosing 3 = 15,600 possible permutations

Digits 0 to 9 = 10 possible choosing 3 = 720 possible permutations

Multiply the two together 15600*720 = 11,232,000 possible permutations<LEGEND></LEGEND>

 

[Nalani]

Digits 0 to 9 = 10 possible choosing 3 = 720 possible permutations

For me, not being a math wiz. But just using the numbers 0-9 wouldn't it come out to 1000. That is 000-999.

I don't see (understand) where you come out with 720

 

[hiker95]

Nalani,

Interesting!!!

How many license plates can be made using three digits and three letters without repetition and alternating letters and digits?

http://wiki.answers.com/Q/How_many_...repetition_and_alternating_letters_and_digits

If it alternates and you start with a letter, then there are 11,232,000 permutations. Then if you start with a number and alternate, there are another 11,232,000 permutations, for a total of 22,464,000 permutations.

If you exclude the letters I, S, B, and O (because they look kind-of like 1, 5, 8, & 0 - kind-of important on license plates) then you are down to 6,652,800 & 13,305,600 respectively.



How many license plates can be made using three digits and three letters without repetitions of letters and digits?

http://wiki.answers.com/Q/How_many_...ers_without_repetitions_of_letters_and_digits

10,108,800 assuming that leading 0s are not allowed in the 3-digit numbers; that the licence authorities do not block some letter combinations because they are rude, and letters that might be confused with numbers (O and 0, I and 1, Z and 2, S and 5, B and 8) are all permitted.

 

[Nalani]

First off, I posted this

for a reason!

Do tell, what is the reason.