There are 11,238,513 of them.
9999
This permutation problem depends on whether the numbers are allowed to be repeated. If they are, there are a possible 9999 numbers, starting with 0001 and running sequentially to 9999. If they are not allowed to be repeated, there are a possible 5040 combinations.
You can make 5 combinations of 1 number, 10 combinations of 2 numbers, 10 combinations of 3 numbers, 5 combinations of 4 numbers, and 1 combinations of 5 number. 31 in all.
First figure in combination can be any of 10, second can be any of remaining 9, third any of remaining 8 and fourth any of remaining 7 so total of possibilities is 10 x 9 x 8 x 7 ie 5040.This assumes no repeating digits, in that case possibilities would be 10 to the fourth power ie 10000.----------------------------------------------------------------------------------------------The question asks for number of combinations. The above result of 5040 is the number of 4 number permutations that can be obtain from the numbers 1 to 10.The number of different 4 number combinations that can be obtain from the numbers1 to 10 is: 10C4 = 10!/(4!∙6!) = 210 different 4 number combinations.If repetition of numbers in the 4 number combination are allowed we have to ad:with 2 numbers repeated; 3∙10C3 = 3(120) = 360with 3 numbers repeated; 2∙10C2 = 2(45) = 90with 4 numbers repeated; 10with two sets of 2 numbers repeated; 10C2 = 45This gives a total of 4 number combinations with repetition of numbers allowed of:210 + 360 + 90 + 10 + 45 = 715 different 4 number combinations.
There are 1,947,792 combinations of 6 numbers out of 36 numbers.
If the numbers can be repeated and the numbers are 0-9 then there are 1000 different combinations.
If numbers may be repeated . . . 36 If numbers may not be repeated . . . 30 If the sequence doesn't matter . . . 15
NOT AN ANSWER:More acceptable combinations:1231 3213 2321More unacceptable combinations:1112 2222 3113Thanks.
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.
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.
How many numbers per combination?
9999
Infinite. If numbers can be repeated, the list could go on nonstop.
9x8x7x6x5x4x3x2x1 or 9! which equals 362880 possible combinations if no digits are repeated
9
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.
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.