given 4 resistors how many different combinations can be made?
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Four resistors yields nine combinations. The advantage is that by using four resistors, the power dissipation capability has also increased by a factor of four, and the voltage withstand doubles. This is a useful trick when you want to keep all your resistors surface-mount types.
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There are about 1,000 combinations possible for a 3-digit number/code. There are 10,000 combinations possible for a 4-digit number/code.
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FACT: most master\lock padlocks have around 64,000 possible combinations. ANOTHER FACT: this instructable ( if done properly) will turn 64,000 into 100.
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However, as we have 4 slots, which need to be all filled at the same time in order to form a 4-digit number, we have to multiply the number of possibilities for each slot together, ie. 2 x 2 x 2 x 2 = = 16. Thus, 16 permutations are made possible.
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Write out a truth table for this circuit. Like the 3-input AND circuit, this circuit will have 8 possible sets of input.
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If repetition of digits is not allowed then the number of combinations possible is 12P12. That is 12! Combinations. So 479001600 combinations are possible if repetition is not allowed.
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-- we're in a position to answer your original question. With 8 items on your menu, the number of possible combinations will be 28 = 256. And if you bought all 8 items, and tried to arrange them in different orders on your tray, the number of arrangements ('permutations') would be 8! = 40320.
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['An Arduino Uno.', 'A HD44780 compatible LCD controller.', 'A 1k ohm resistor.', 'A 10k ohm resistor.', 'A 220 ohm resistor.', 'A breadboard and a bunch of jumper cables.']
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Answer: There are 9,999 different combinations in a 4-digit lock.
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There are four possible combinations. All three in series. All three in parallel.
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It's the same as possibilities to a 6 digit code with known numbers and no repeats. With 1 item there are 6 possibilities with 2 you only have 5 choices after the first if you don't repeat the same item. So for 6 items the equation is as follows 6*5*4*3*2= 720 possible combinations of 6 items.
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reto wrote: How many different 4 letter combinations can be created from the letters A,B,C, and D? This is a case of number of arrangements possible with repetition. Thus you have 4 options each for the 1st,2nd,3rd and 4th letter giving you a total of 4*4*4*4 = 256 number of possible combinations.
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If you allow repetitions of any digit, then there are 4 choices for each of the 3 positions in the 3-digit numbers, or 4 x 4 x 4 = 64 3-digit numbers, in total.
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There are 10,000 possible combinations that the digits 0-9 can be arranged into to form a four-digit code.
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But if you want to know how many different license plates you can make with six characters, assuming that you allow letters and numbers (that's 36 characters per digit), the formula for calculating the number of possible combinations is 36 6, which comes out to 2,176,782,336.
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They should all be the same. There's not a lot of room for variation with only two wires.
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Dana F Anderson, Just interested in math is all. Kinda a hobby. Assuming the digits 0 through 9 can be used, and a number can be repeated in the combination, then there are 10^3 possible combinations (1000 - from 000 to 999). If three different numbers have to be used, then 10 x 9 x 8 = 720 possible combinations.