

Subject:
Math problem (Combinations)
Category: Science Asked by: mrsneakyga 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 answer?  
 
 


Subject:
Re: Math problem (Combinations)
Answered By: dogbitega on 30 Apr 2003 09:53 PDT Rated: 
Hi mrsneakyga, 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. dogbitega  
 

mrsneakyga rated this answer: 

Subject:
Re: Math problem (Combinations)
From: research_helpga 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: secret901ga 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. secret901ga 
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