1-36 are numbers, NOT digits.
The number of combinations of 10 numbers out of 36 is
36*35*34*33*32*31*30*29*28*27/(10*9*8*7*6*5*4*3*2*1)
which is 254,186,856
If the digits can repeat, then there are 256 possible combinations. If they can't repeat, then there are 24 possibilities.
If the numbers contain zeros, the total number of combinations is 10,000. You can work this out easily logically: For ten single-digit numbers (0,1,2,3,4,5,6,7,8,9) then there are 10 possible 'combinations' For numbers with 2 digits then for each possible digit in the 10s column (e.g. in the 20s range) there are another 10 possible combinations (20,21,22,23,24,25,26,27, 28,29). As there are 10 possible ranges (single digits, teens, twenties, thirties etc) there will be 10 X 10 or 100 possible combinations. using the same logic, for three digits, there will be 10 X 10 X 10 or 1000 digits. And for 4 digits there will be 10 x 10 x 10 x 10 = 10,000 possible combinations. So for a number, say, with x digits, the total number of combinations of those digits will be 10 x 10 x 10..... etc with x numbers of 10s in the calculation. You can find out the number of combinations of any set of letters or numbers in the same way. as an example, to find out, say, the possible combinations of letters in the alphabet of 26 letters, then using the same method this can be given as 26 x 26 x 26 x 26............. with 26 '26's' in a row multiplied together. This gives the staggering amount of approximately 615612 followed by 31 zeros.
To calculate the number of 4-digit combinations that can be made with 4 digits, we can use the formula for permutations. Since there are 10 possible digits (0-9) for each of the 4 positions, the total number of combinations is 10^4, which equals 10,000. This is because each digit can be selected independently for each position, resulting in a total of 10 choices for each of the 4 positions.
Total of 720 combinations:1 --- 12 --- 23 --- 64 --- 245 --- 1206 --- 7207 --- 5040.......Notice that:A --- BC --- DE --- FG --- XIn this sequence, if you wanted to know "X", it would be equal to FxG, and that is the key to this sequence. Why does it happen?Well, if, lets see, for instance, if 2 digits (0 and 1) have a possible of two combinations (10 and 01), then 3 digits (0, 1 and 2) will have a possible of six combinations: 012021And equivelent, but switching 0 with either 1 or 2, making it repeat itself 3 times.Eventually, we will end up doing 2 (number of combinations with 2 digits) times 3 (number of digits of which we want to know the number of unique combinations possible), totallizing 6 (our wanted-to-know number)
2087
If the digits can repeat, then there are 256 possible combinations. If they can't repeat, then there are 24 possibilities.
If the numbers contain zeros, the total number of combinations is 10,000. You can work this out easily logically: For ten single-digit numbers (0,1,2,3,4,5,6,7,8,9) then there are 10 possible 'combinations' For numbers with 2 digits then for each possible digit in the 10s column (e.g. in the 20s range) there are another 10 possible combinations (20,21,22,23,24,25,26,27, 28,29). As there are 10 possible ranges (single digits, teens, twenties, thirties etc) there will be 10 X 10 or 100 possible combinations. using the same logic, for three digits, there will be 10 X 10 X 10 or 1000 digits. And for 4 digits there will be 10 x 10 x 10 x 10 = 10,000 possible combinations. So for a number, say, with x digits, the total number of combinations of those digits will be 10 x 10 x 10..... etc with x numbers of 10s in the calculation. You can find out the number of combinations of any set of letters or numbers in the same way. as an example, to find out, say, the possible combinations of letters in the alphabet of 26 letters, then using the same method this can be given as 26 x 26 x 26 x 26............. with 26 '26's' in a row multiplied together. This gives the staggering amount of approximately 615612 followed by 31 zeros.
*inclusive: 74 = 2401 *exclusive: 7*6*5*4 = 840
To find the number of 3-digit combinations using the digits 0 to 9 with repetition allowed, we consider that each digit can be any of the 10 digits (0-9). Since there are 3 positions in the combination, the total number of combinations is calculated as (10 \times 10 \times 10), which equals 1,000. Therefore, there are 1,000 possible 3-digit combinations.
To calculate the number of 4-digit combinations that can be made with 4 digits, we can use the formula for permutations. Since there are 10 possible digits (0-9) for each of the 4 positions, the total number of combinations is 10^4, which equals 10,000. This is because each digit can be selected independently for each position, resulting in a total of 10 choices for each of the 4 positions.
To find the number of possible combinations of the digits in the number 7336, we need to consider the arrangement of the digits while accounting for the repeating digit '3'. The formula for permutations of a multiset is given by ( \frac{n!}{n_1! \times n_2! \times ...} ), where ( n ) is the total number of items and ( n_1, n_2, ) etc., are the frequencies of the repeating items. Here, the total number of digits is 4 (7, 3, 3, 6), so the calculation is ( \frac{4!}{2!} = \frac{24}{2} = 12 ). Thus, there are 12 unique combinations of the digits in 7336.
There are 1,000,000 possible 6-digit combinations. This is calculated by considering that each digit can range from 0 to 9 (10 possible values), and since there are 6 digits, the total combinations are (10^6 = 1,000,000).
The number 1379 can be expressed as a combination of its digits in various ways. The individual digits can be rearranged, yielding combinations such as 1379, 1397, 1739, 1793, 1937, 1973, 3179, 3197, 3719, and so on. Additionally, combinations can also refer to grouping the digits, such as 1, 3, 7, and 9, or pairs like (1, 3), (1, 7), (1, 9), etc. The total number of unique arrangements of the digits, however, is 4! = 24 different permutations.
Assuming the digits cannot be repeated, there are 7 combinations with 1 digit, 21 combinations with 2 digits, 35 combinations with 3 digits, 35 combinations with 4 digits, 21 combinations with 5 digits, 7 combinations with 6 digits and 1 combinations with 7 digits. That makes a total of 2^7 - 1 = 127: too many for me to list. If digits can be repeated, there are infinitely many combinations.
The total number of 1-bit combinations is 2. This is because a single bit can have two possible values: 0 or 1. Therefore, the combinations are {0, 1}.
Total of 720 combinations:1 --- 12 --- 23 --- 64 --- 245 --- 1206 --- 7207 --- 5040.......Notice that:A --- BC --- DE --- FG --- XIn this sequence, if you wanted to know "X", it would be equal to FxG, and that is the key to this sequence. Why does it happen?Well, if, lets see, for instance, if 2 digits (0 and 1) have a possible of two combinations (10 and 01), then 3 digits (0, 1 and 2) will have a possible of six combinations: 012021And equivelent, but switching 0 with either 1 or 2, making it repeat itself 3 times.Eventually, we will end up doing 2 (number of combinations with 2 digits) times 3 (number of digits of which we want to know the number of unique combinations possible), totallizing 6 (our wanted-to-know number)
The 120 combinations for the digits 1, 2, 3, 4, and 5 refer to the permutations of these five distinct digits. Each arrangement of these digits counts as a unique combination, and since there are 5 digits, the total number of permutations is calculated as 5! (5 factorial), which equals 120. Examples include 12345, 12354, 12435, and so on, continuing through all possible arrangements of these five numbers.