If you allow for zeros, that's 10^16, which is ten quintillion, or 10,000,000,000,000,000.
If not, it's 9^16, which is 1,853,020,188,851,841, or 1.85 quintillion. (1.85x10^15)
To calculate the number of 4-digit combinations you can get from the numbers 1, 2, 2, and 6, we need to consider that the number 2 is repeated. Therefore, the total number of combinations is calculated using the formula for permutations of a multiset, which is 4! / (2!1!1!) = 12. So, there are 12 unique 4-digit combinations that can be formed from the numbers 1, 2, 2, and 6.
There are 27 of them.
There are 167960 9 digits combinations between numbers 1 and 20.
90
If numbers can be repeated and zeroes are allowed to lead, this is simply all natural numbers in the set {0000 - 9999}, for a total of 10,000 possible combinations. If numbers cannot be repeated, this becomes a permutation problem; out of 10 possible digits, permute four of them. This evaluates as: nPr = n! / (n-r)! 10P4 = 10! / (10-4)! 10P4 = 1 * 2 * 3 * ... * 6 * 7 * 8 * 9 * 10 / 1 * 2 * 3 * 4 * 5 * 6 10P4 = 7 * 8 * 9 * 10 10P4 = 5040 There are thus 5,040 possible combinations of four of the digits in 0-9 if any digit is not used twice and 0 is allowed to be a leading digit.
Assuming that repeated numbers are allowed, the number of possible combinations is given by 40 * 40 * 40 = 64000.If repeated numbers are not allowed, the number of possible combinations is given by 40 * 39 * 38 = 59280.
The number of four-digit combinations is 10,000 .Stick a '3' before each of them, and you have all the possible 5-digit combinations that start with 3.There are 10,000 of them. They run from 30,000 to 39,999 .
720 (10*9*8)
9x8x7x6x5x4x3x2x1 or 9! which equals 362880 possible combinations if no digits are repeated
9
Five-digit combinations of numbers from 0 to 9 can range from 00000 to 99999. Each digit can independently be any number from 0 to 9, leading to a total of 10 options per digit. Consequently, there are (10^5) or 100,000 possible combinations. Each combination can include repeated digits and leading zeros.
To calculate the number of 4-digit combinations you can get from the numbers 1, 2, 2, and 6, we need to consider that the number 2 is repeated. Therefore, the total number of combinations is calculated using the formula for permutations of a multiset, which is 4! / (2!1!1!) = 12. So, there are 12 unique 4-digit combinations that can be formed from the numbers 1, 2, 2, and 6.
The answer will depend on how many digits there are in each of the 30 numbers. If the 30 numbers are all 6-digit numbers then the answer is NONE! If the 30 numbers are the first 30 counting numbers then there are 126 combinations of five 1-digit numbers, 1764 combinations of three 1-digit numbers and one 2-digit number, and 1710 combinations of one 1-digit number and two 2-digit numbers. That makes a total of 3600 5-digit combinations.
Number of 7 digit combinations out of the 10 one-digit numbers = 120.
NOT AN ANSWER:More acceptable combinations:1231 3213 2321More unacceptable combinations:1112 2222 3113Thanks.
If you allow digits to be repeated (for example, 222 or 992), then there are 9 x 9 x 9 = 729 combinations. If you do not allow digits to be repeated, then there are 9 x 8 x 7 = 504 combinations.
Using the digits of 1345678, there are 210 three digit numbers in which no digit is repeated.