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 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 answer?``` 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? secret901-ga``` 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```
 Subject: Re: Math problem (Combinations) Answered By: dogbite-ga on 30 Apr 2003 09:53 PDT Rated:
 ```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``` 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``` 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```