Twenty-four. You have four choices for the first digit. Once selected, you have three choices for the second digit. Then two for the third and then the last digit.
Multiply 4 X 3 X 2 X 1 and you get 24, which is 4! or "4 factorial".
There are 720 permutations of 3 different digits from the ten digits 0 to 9, including leading zeros.
This is permutations with repetition. The answer is 4^4 = 256 total permutations. Since 2 of the digits used are odd (and 2 are even), then half of the possibilities will be odd: 128 odd numbers.
The answer is 26*26*26*10*10*10 = 17,576,000
If the 6 digits can be repeated, there are 1296 different combinations. If you cannot repeat digits in the combination there are 360 different combinations. * * * * * No. That is the number of PERMUTATIONS, not COMBINATIONS. If you have 6 different digits, you can make only 15 4-digit combinations from them.
If you wish to use a brute-force method, assuming the digits go from 0 to 9 there are 10,000 possible permutations.
How many permutations of 3 different digits are there, chosen from the ten digits 0 to 9 inclusive?
Assuming you mean permutations of three digits, then the set of numbers that can be made with these digits is: 345 354 435 453 534 543 There are six possible permutations of three numbers.
There are 720 permutations of 3 different digits from the ten digits 0 to 9, including leading zeros.
8 digits will generate over 40,000 permutations.
This is permutations with repetition. The answer is 4^4 = 256 total permutations. Since 2 of the digits used are odd (and 2 are even), then half of the possibilities will be odd: 128 odd numbers.
6 of them. 4C2 = 4!/(2!*2!) = 4*3/(2*1) = 6
In case of digits or numbers, when we say combination, since the order of the numbers matters, we really need to find the permutations. For example, 12 and 21 are two permutations and not combination. The permutation for 10 digits chosen 5 ways is 10*9*8*7*6, which is 30240. But how many are there in whole like four digit has 24.
It is possible to create infinitely many numbers, of infinitely many different lengths, using the digits of the given number. Using each of the digits, and only once, there are 5! = 120 different permutations.
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.
The answer is 26*26*26*10*10*10 = 17,576,000
If the 6 digits can be repeated, there are 1296 different combinations. If you cannot repeat digits in the combination there are 360 different combinations. * * * * * No. That is the number of PERMUTATIONS, not COMBINATIONS. If you have 6 different digits, you can make only 15 4-digit combinations from them.
There is only one combination. The order of the digits in combinations makes no difference. They are considered as being different if they are permutations, not combinations.