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I'm working with an electrical enginier and we are consturcting a control panal, we have a item that we are controling with 4 diffent swithes that each have 3 positions. We were wondering how many possibilites there are for someone to mess these controls up. Could some one please help us and give us the formula for figuring this out. THANK YOU

2006-08-10 10:44:03 · 13 answers · asked by Jim S 1 in Science & Mathematics Mathematics

13 answers

81 different positions. 3(positions)^4(controls)

2006-08-10 10:49:16 · answer #1 · answered by shmifty__14 5 · 1 4

3 x 3 x 3 x 3 = 81

2006-08-10 10:49:29 · answer #2 · answered by Paul 3 · 0 0

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 · answer #3 · answered by Anonymous · 0 0

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 · answer #4 · answered by Will 6 · 0 0

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 · answer #5 · answered by MollyMAM 6 · 0 0

3 positions for 4 switches would be 3x3x3x3 = 81

2006-08-10 10:50:47 · answer #6 · answered by David S 2 · 0 0

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 · answer #7 · answered by Anonymous · 0 0

4 x 3 = 12
swiches positions possibilities

2006-08-10 11:54:47 · answer #8 · answered by BENNY C 2 · 0 0

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 · answer #9 · answered by sandwreckoner 4 · 0 0

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 · answer #10 · answered by a_liberal_economist 3 · 0 0

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 · answer #11 · answered by shclapitz 3 · 0 0

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