81 different positions. 3(positions)^4(controls)
2006-08-10 10:49:16
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answer #1
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answered by shmifty__14 5
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3 x 3 x 3 x 3 = 81
2006-08-10 10:49:29
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answer #2
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answered by Paul 3
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There is one possible correct combination?
Multiply each switch's number of positions by the number of each other switches' positions:
3*3*3*3
=81 possible positions
Number of incorrect positions = 81 - 1 = 80
2006-08-10 13:54:01
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answer #3
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answered by Anonymous
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This is essentially a 4 digit number in a trinary number system (3 possible digits). The number of possible combinations range from 0000 to 2222 (base 3). So there are 10000(base 3) combinations. Converted to decimal its 3^4 = 81.
2006-08-10 10:50:40
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answer #4
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answered by Will 6
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It's called the Counting Principle - you multiply together the separate number of possibilities:3*3*3*3 =81
2006-08-10 10:53:31
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answer #5
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answered by MollyMAM 6
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3 positions for 4 switches would be 3x3x3x3 = 81
2006-08-10 10:50:47
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answer #6
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answered by David S 2
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In general if you have k switches with n possible values for each switch then the number of combinations is
n ^ k
In your case there are 4 switches and 3 possible values
3 ^ 4 = 81
If there are x possible correct states then there are
n ^ k - x
ways to screw it up although i presume how badly would depend on which combination
2006-08-10 10:52:22
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answer #7
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answered by Anonymous
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4 x 3 = 12
swiches positions possibilities
2006-08-10 11:54:47
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answer #8
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answered by BENNY C 2
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S(3)*S(3)*S(3)*S(3) = 3^3 or (S)^3, There are 81 different combos. (S) stands for switch.
2006-08-10 12:32:16
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answer #9
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answered by sandwreckoner 4
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Engineers usually take a fair amount of math, so (s)he should be able to answer the question...that is, if there was one.
Nothing like Yahoo Answers to get your homework done...
2006-08-10 10:53:30
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answer #10
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answered by a_liberal_economist 3
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I would like to be the devils advocate...2x3x3x3=54 why? becauce one is the on and off switch.
2006-08-10 11:04:24
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answer #11
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answered by shclapitz 3
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