L O.

asked • 05/23/14

help with possibilities for a four digit number

suppose a contest offers $1,000 to the person who can guesss the winning four digit number.  How do i solve to find how many possibilities?
 
 
thanks.

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Jim S. answered • 05/23/14

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Hi,
      You have 10 choices for the first number, 10 choices for the second and so on, so there are 10*10*10*10 or 10,000 4 digit numbers.
 
Jim
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Bob A.

But you are saying that 1 is a four digit number 0001 and should be counted.
Not to disagreewith that, but this, to me, is a question that could be interpreted two ways.
Should we be counting 0 to 9,999 for 10,000  four digit numbers?
Or should we only be counting 1,000 to 9,999, and not counting 0 to 999 - all those 1, 2, and 3 digit numbers?
I would say this question is poorly written.  Not to specify if 1 is a four digit number to be counted or not; and also what has the $1,000 contest got to do with anything.
All a consequence of asking a math question using the very imprecise English language.
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05/23/14

Jim S.

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Robert,
         Thanks for your comments. Hey, 0001 is a 4 digit number, if it's a lottery (hence the $1000) and the 4 numbers are drawn with replacement then 0001 is just as likely as 2457 And there are 10,000 of them.
       Many of the questions have multiple interpretations, I try an pick one.
Regards
Jim
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05/23/14

L O.

I do have a question.  I tried solving this problem into mutually exclusive cases.
 
9x4x4x4 =       576
9
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05/24/14

L O.

I have a question.  I tried solving this using mutually exclusive cases.  Did I do this correctly?
 
10 x 4 x 4 x 4 =              640
10 x 10 x 4 x 4 =          1600
10 x 10 x 10 x 4 =        4000
10 x 10 x 10 x 10 =    10000
 
                                ________
 
                                  16240
          
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05/24/14

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