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 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.
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.
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)
Simple enough to solve. The answer is a power of two. Assuming you have two possible digits, say for example, 3 and 4, then you simply have to multiply it by how many numbers you want to get the total number of combinations. Each number can be 3 or 4 in this case, and you have 5 numbers. That's two to the fifth. Five combinations of any two numbers. 2x2x2x2x2. The answer is 32 combinations.
There are 10 digits from which the first number can be selected, 10 digits from which the second number can be chosen and 10 from which the third number can be picked. This gives you a total number of combinations of 10 x 10 x 10 = 1000.
2087
As the number has to start with 15, we have only 3 remaining digits to work with. There are 3 possible options for the first digit. Then out of each of these, 2 possible options for the second digit, and one option for the last. This means that in total there are 3x2x1 (6) possible combinations. These are: 15234 15243 15324 15342 15423 15432
To calculate the number of 4-digit combinations using the digits 1, 3, 5, and 7 exactly once each, we can use the permutation formula. There are 4 choices for the first digit, 3 choices for the second digit, 2 choices for the third digit, and 1 choice for the fourth digit. Therefore, the total number of combinations is 4 x 3 x 2 x 1 = 24. So, there are 24 possible 4-digit combinations using the digits 1, 3, 5, and 7 exactly once each.
To find the number of three-digit combinations, we consider the digits from 000 to 999. Each digit can range from 0 to 9, giving us 10 options for each of the three digits. Therefore, the total number of three-digit combinations is (10 \times 10 \times 10 = 1,000).