There are 126 different 5 digit combinations.
Note that the combination 12345 is the same as the combination 45312.
The question can be re-stated as asking for the total number of permutations that can be derived from the following two groups of digits: AAABCD and AABBCD, where A, B, C and D are different. The number of ways of choosing the digit A, to be used three times, out of the ten digits {0, 1, 2, ... 9} is 10. Having done that, the number of ways of selecting 3 from the remaining 9 digits is 9C3 = (9*8*7)/(3*2*1) = 84. Thus there are 10*84 = 840 combinations of the form AAABCD. You now need all the distinct permutations of these 6 digits. The total number of permutations of 6 digits is 6! but because 3 of the digits are the same, these permutations are not all distinct. In fact, there are 6!/3! = 120 distinct permutations. That makes a total of 840*120 = 100800 such numbers. Next, the number of ways of choosing the digits A and B, each to be used twice, out of the ten digits {0, 1, 2, ... 9} is 10C2 = (10*9)/(2*1) = 45. Having done that, the number of ways of selecting C and D from the remaining 8 digits is 8C2 = (8*7)/(2*1) = 28. Thus there are 45*28 = 1260 combinations of the form AAABCD. You now need all the distinct permutations of these 6 digits. The total number of permutations of 6 digits is 6! but because 2 pairs of these digits are the same, these permutations are not all distinct. In fact, there are 6!/(2!*2!) = 180 distinct permutations. That makes a total of 1260*180 = 226800 such numbers. The grand total is, therefore, 100800 + 226800 = 327600 6-digit numbers made from 4 distinct digits.
There is only one five-number combination of 5 distinct numbers.
To calculate the number of combinations for the numbers 1248, we need to consider all possible arrangements of the four digits. Since all the digits are unique, there are 4 factorial (4!) ways to arrange them. This equals 4 x 3 x 2 x 1 = 24 combinations.
With repeating digits, there are 33 = 27 possible combinations.Without repeating any digits, there are 6 combinations:357375537573735753
Actually the product of a nonzero rational number and another rational number will always be rational.The product of a nonzero rational number and an IRrational number will always be irrational. (You have to include the "nonzero" caveat because zero times an irrational number is zero, which is rational)
There are infinite combinations that can make 3879
Two. All nonzero digits are significant.
11
There are different numbers of combinations for groups of different sizes out of 9: 1 combination of 9 digits 9 combinations of 1 digit and of 8 digits 36 combinations of 2 digits and of 7 digits 84 combinations of 3 digits and of 6 digits 126 combinations of 4 digits and of 5 digits 255 combinations in all.
The number of distinct combinations that can be created with n bits is 2n.
Four. All nonzero digits are significant.
Five. All nonzero digits are significant.
A number with nonzero digits after the decimal points is not a whole number, and it isn't equal to any whole number.
Since a number can have infinitely many digits, there are infinitely many possible combinations.
128
The number 48 cm has two significant digits. The leading 0 is not significant in this case.
You Can Create 999 Number combinations