how many different combinations in a rubik's cube?
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The original 3x3x3 Rubik's cube has 43 252 003 274 489 856 000 combinations, or 43 quintillion. Again, as pointed out on this website's main page, this is a manageably imaginable number. It's a little less than the square of the earth's population, for example.
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Combinations are selections of objects, with or without repetition, order does not matter. The number of k-element combinations of n objects, without repetition is Cn,k = (n k ) = n! k!( n − k)! .
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There are 325 possible combinations with two letters. To determine this number of combinations, we use the fact that the alphabet has 26 letters....
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We conclude that there are 35−3×25+3=150 combinations. If there were no restrictions, the number of five digit codes that could be formed with the digits 1, 2, 3 would be 35 since there are three ways to choose each of the five digits.
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The Rubik's cube (sometimes misspelled rubix cube) is a mechanical 3D puzzle, invented more than 30 years ago and still considered as the best-selling toy of all times! Yet, solving the Rubik's Cube is considered a nearly-impossible task, which requires an IQ of 160...
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Given 100, choose 98, there are 161,700 combinations. The number of combinations gets bigger fast. 100, choose 95, is over 7 million combinations. The largest number of combinations given a set of 100 elements, will be 100, choose 50.
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There are 40,475,358 possible 6-number combinations that can be formed from numbers 1-58 . To determine this, the Combination formula is used. n choose k equals the factorial of n (divided by) all over the factorial of k times (or multiplied by) the factorial of n minus k.
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The number of combinations of n objects taken r at a time is determined by the following formula: C(n,r)=n! (n−r)! r!
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There are, you see, 3 x 2 x 1 = 6 possible ways of arranging the three digits. Therefore in that set of 720 possibilities, each unique combination of three digits is represented 6 times.
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Answer: There are 9,999 different combinations in a 4-digit lock.
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It is the number of ways to choose 3 of the 10 digits in which order is not important. There are 6 = 3x2x1 ways to order 3 digits in a row. Thus the number of combinations of 3 of the 10 digits is 720/6 = 120 combinations.
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There are 52 cards in each deck, meaning that the total possible combinations would equal 52! (52 factorial) which is equal to 8.06e+67. This number is MASSIVE.
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Assuming no five-digit number can begin with zero, there are 9 possible choices for the first digit. Then there are 10 possible choices for each of the remaining four digits. Therefore, you have 9 x 10 x 10 x 10 x 10 combinations, or 9 x 10^4, which is 90,000 different combinations.
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Since each friend can either attend or not, there are 2x2x2x2x2x2 = 2^6 = 64 different combinations, but that includes one where none of the friends attends, so subtract 1 for that instance, for a total of 63 combinations. 64–1= 63.
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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.
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There are 60 of them. Using the same argument as above there are 60 60 = 3600 two "digit" combinations and 60 60 60 = 216000 three "digit" combinations.
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Seven of the corners can be independently rotated, and the orientation of the eighth depends on the other seven, giving 8! ×37 combinations. There are 144 movable centers, consisting of six sets of 24 pieces each. Within each set there are four centers of each color.