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Combinations of Letters and Numbers


Date: 3/19/96 at 10:22:57
From: Mike Hansen
Subject: Math Contests

I just participated in a Math Contest. I missed 5th place by 1 
point. I had a question about a problem in it:

A moped license plate has two letters and then four numbers in it. 
How many plates can be made without duplicating?

Mike Hansen


Date: 3/20/96 at 1:35:8
From: Doctor Patrick
Subject: Re: Math Contests

Hi Mike!  Is that  without duplicating numbers and letters, or 
plates?  

Since I'm not sure, I'll show you how to do both.  

The easier one is to find is the second, so let's start with 
that. The first digit is a letter, so you have 26 choices.   Each 
of those 26 possible choices for the first digits can be followed 
by 26 more choices for the second digit.  So to find the total 
number of letter combinations we multiply the number of choices 
for the first digit by the number of choices for the next digit, 
in this case 26*26, for a total of 676.

Do you see how that works?

Then we do the ame for the numbers. There are four digits and 
each of them can be from 0 to 9, for 10 possible choices per 
digit. That gives us 10*10*10*10, 10000 combinations.  

Now we just need to combine the two parts.  There were 676 
possible letter combinations  multiplied by 10000 number 
combinations for each of them.  I'll let you calculate the final 
step on your own, using the methods from above.

In the case of no duplicate digits we must multiply the 26 
possible first digits by 25 (26 letters - the one already used.  
This gives us 26*25 letters, which equals 650.  Then we move on to 
the numbers.  There are 10 for the first digit, then 9 for the 
second (10 -1 for the number already used).  

How many do you think will be in the third and fourth digits? 

Once you have those four numbers, multiply them out as we did 
above, and then put the two parts (letters and numbers) together.

Good luck!

-Doctor Patrick,  The Math Forum


Date: 3/20/96 at 9:54:1
From: Mike Hansen
Subject: Re: Math Contests

I got: 

26(letters) * 26(letters) * 10(digits) * 10(digits) * 10(digits) *
10(digits) = 6,760,000

That is what they had on the answer sheet, but they changed it to 
three million something. My teacher and I do not know how they 
came up with this answer. My way is perfectly logical.


Date: 3/20/96 at 11:42:24
From: Doctor Patrick
Subject: Re: Math Contests

Was the answer they gave 3,276,000?  If so, then they wanted you 
to do the problem in the second way I gave.  This was the number 
of plates that could be found with no letters or numbers repeated 
inside of the same plate.  

26 (letters) * 25 (the unused letters) *10 (numbers) * 9 (unused 
numbers) ...

Do you think you can finish this on your own?  First find the 
number of remaining choices for the last two digits, then multiply 
it out.

Hope this helps,

-Doctor Patrick,  The Math Forum

    
Associated Topics:
High School Permutations and Combinations

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