The number of combinations for 9 digits can be calculated using the formula for permutations of n objects taken r at a time, which is n! / (n-r)!. In this case, with 9 digits, there are 9! / (9-9)! = 9! / 0! = 9! = 362,880 possible combinations. This is because each digit can be arranged in 9 different positions, and as each digit is unique, the total number of combinations is the product of these possibilities for each digit.
There are only 10 combinations. In each combination one of the 10 digits is left out.
9
10000
There are 10 digits, 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9. Let's say that a three digits number has a form of xyz. There are only 9 possibilities for x (0 cannot be in the first place), 10 possibilities for y, and 10 possibilities for z. So there are 9 x 10 x 10 = 900 combinations. However, there are only 9 digits from 1 to 9. So we have 9 possibilities for x, y, and z. Hence, there are 9 x 9 x 9 = 729 combinations.
To calculate the number of possible combinations of the digits 1, 3, 7, and 9, we can use the formula for permutations of a set of objects, which is n! / (n-r)!. In this case, there are 4 digits and we want to find all possible 4-digit combinations, so n=4 and r=4. Therefore, the number of possible combinations is 4! / (4-4)! = 4! / 0! = 4 x 3 x 2 x 1 = 24. So, there are 24 possible combinations using the digits 1, 3, 7, and 9.
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.
There are only 10 combinations. In each combination one of the 10 digits is left out.
9
There are 167960 9 digits combinations between numbers 1 and 20.
Since 0 cannot be in the first place of the 4-digits number, we have 9 x 10 x 10 x 10 = 9,000 combinations.
1000
Perhapsf you specified the number of digits. e.g. 0-9 with two digits.
0000-9999 (10x10x10x10 or 104) = 10,000 possible combinations allowing for repeated digits. If you are not able to repeat digits then it's 10 x 9 x 8 x 7 or 5,040 possible combinations without repeated digits.
9.
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That depends on how many of the digits are repeated. If no digits are repeated, you have 9 choices for the first digit, 8 choices for the second, etc.; for a total of 9 x 8 x 7 x 6 x 5 x 4.
45 In combinations, the order of the digits does not matter so that 12 and 21 are considered the same.