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**roel****Member**- Registered: 2006-07-22
- Posts: 48

can u please help me with this? Here's the problem:

car plates here in the philippines consists of 3 letters and 3 numbers... letters are from A-Z and numbers are from 0-9.. (For example, ABC-123).. The question is how many permutations, arrangement, or combinations can we get? the Letters referred may start in any letter but on part of the digits, it must start only from 1 to 9 and not 0. each letter and number must appear in each permutation only once. Take note that in each permutation, letters and numbers must be combined. Therefore, e.g. ABC-123, ABC-124, ABC-125 are considered distinct from one another.. Actually, this is our assignment. I already made it but i'm not sure with my answer. That's why i'm asking you experts..

It is already 11:51 in the evening here guys.. and i only have until 11:30 in the morning to check my e-mail and read the answer. Please help me with these... Thanks a lot..

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**pi man****Member**- Registered: 2006-07-06
- Posts: 251

You have 26 * 25 * 24 combinations of letters you can use assuming no duplicates. That's 15600. You can choose from 1 to 9 for the 1st digit. For the second digit you have 9 choices also; 0 through 9 except for whatever was picked for the 1st digit. The 3 digit you have 8 choices. So 9*9*8 possibilities for the numeric part of your plate. That's 648. Multiply those two numbers together: 648 * 15600 = 10,108,800.

I'm not sure what you mean by "letters and numbers must be combined". Do you mean the letters must be grouped together and the numbers must be grouped together? ABC123 is valid but AB123C is not? If so, then the only 2 formats possible are AAA### and ###AAA. In that case multiply the 10108800 by 2 for the total possibilities.

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**roel****Member**- Registered: 2006-07-22
- Posts: 48

yes.. they must be grouped together... thanks a lot

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**brianN****Guest**

I think the possibilities of 3-lettered group are 26x26x26=17576, of 3-digitted group are 10x10x10=1000, but without 0 being the first number (000-099), we have 1000-100=900 possibilities. Multiply them together, we have 17576x900=15818400 possibilities for one pattern ABC123 or 123ABC. If both patterns are applied, multiply it by 2.

**brianN****Guest**

roel wrote:

each letter and number must appear in each permutation only once.

I miss this line. Never mind my post above.

**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,651

Not a problem, brian ... (join the forum!)

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman

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