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Q: Math problem (Combinations) ( Answered 5 out of 5 stars,   2 Comments )
Subject: Math problem (Combinations)
Category: Science
Asked by: mrsneaky-ga
List Price: $4.00
Posted: 29 Apr 2003 20:42 PDT
Expires: 29 May 2003 20:42 PDT
Question ID: 197301
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

Clarification of Question by mrsneaky-ga on 29 Apr 2003 21:57 PDT
insert comma after 45,000.   "45,000, 90,000"

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?


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,
Subject: Re: Math problem (Combinations)
Answered By: dogbite-ga on 30 Apr 2003 09:53 PDT
Rated:5 out of 5 stars
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


        16 ^ 180,000


  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.


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.


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


  Also, my symbolic computer gives:

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

  I hope that clears everything up.

mrsneaky-ga rated this answer:5 out of 5 stars

Subject: Re: Math problem (Combinations)
From: research_help-ga on 30 Apr 2003 10:16 PDT
Dogbite - How can you be confident in your answer without the asker
posting the clarifications posted by the other 2 researchers?
Subject: Re: Math problem (Combinations)
From: secret901-ga on 30 Apr 2003 12:50 PDT
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.


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