If a chair can be 1 of 16 colors and there are 45,000 90,000 and
180,000 places to put chairs (3 seperate math problems).  Picking
randomly and equal chance to pick a color chair.  How many possible
combinations are possible?  Please show work.  How many digits in the
answer?

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Clarification of Question by mrsneaky-ga on 29 Apr 2003 21:57 PDT

insert comma after 45,000.   "45,000, 90,000"

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Request for Question Clarification by secret901-ga on 29 Apr 2003 22:02 PDT

Hi mrsneaky,
Are we assuming that there is an unlimited number of chairs for each color?

secret901-ga

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Request for Question Clarification by elmarto-ga on 30 Apr 2003 09:05 PDT

Do all combinations where you have, say, 1 blue and 44,999 red count
as one combination? Or would you rather count as being different
combinations to have the first chair in blue and all the other in red,
and to have the second one in blue and the other ones in red?

Best regards,
elmarto

Hi mrsneaky-ga,

  Suppose we have 45,000 spots for chairs.
  Each spot can hold a chair with 1 of 16
  colors.  That means you can have:

        16 ^ 45,000

  possible arrangements.  Similarly, with
  90,000 and 180,000 spots you have

        16 ^ 90,000

  and

        16 ^ 180,000

  respectively.

  To compute the number of digits, let
  'y' be 45,000, 90,000, or 180,000 and
  'x' be the number of digits in the result,
  and solve

            16^y  =  10^x
     log_16(16^y) =  log_16(10^x)
                y =  x * log_16(10)
                x =  y / log_16(10)
                x =  y / (log_10(10) / log_10(16) )
                x =  y / .83  (approximately)

  Then you have, approximately:

       y       x
    45,000     54,217
    90,000    108,433
   180,000    216,867

  I hope that helps you.

                 dogbite-ga

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Clarification of Answer by dogbite-ga on 30 Apr 2003 14:18 PDT

I'm responding to comments.

    1) I did not wait for a response because
       I thought the origional question was
       clear enough.

    2) The digit count is approximate.  mrsneaky
       wants an answer that uses math, not
       one that counts digits from a symbolic
       evaluator.  Take the ceiling of 1.2
       to get 2 digits. And 31 digits out of 
       over 54,000 is a small error.

                   dogbite-ga

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Clarification of Answer by dogbite-ga on 30 Apr 2003 14:49 PDT

In case you do want exact answers
  for the number of digits, either
  use more precision than 0.83, such as

0.83048202372184058696

  Also, my symbolic computer gives:

$ for i in 45000 90000 180000; do dc -e "16 $i ^ p" | tr -d '\n\\' | wc -c; done
  54186
 108371
 216742

  I hope that clears everything up.

            dogbite-ga

Dogbite - How can you be confident in your answer without the asker
posting the clarifications posted by the other 2 researchers?

If we use the answerer's formula for x, then we would conclude that 16
has 1 digit, viz.:
16^1 = 10^x
1 = x log_16(10)
x = 1/.83 = 1.2
which is obviously wrong.

For example, 16^45,000 has exactly 54,186 digits, which has an error
of 31 from the answerer's approximation.

secret901-ga