Using the formula n!/r!(n-r)! where n is the number of possible numbers and r is the number of numbers chosen, there are 13983816 combinations of six numbers between 1 and 49 inclusive.
If there are no restrictions on duplicated numbers or other patterns of numbers then there are 10 ways of selecting the first digit and also 10 ways of selecting the second digit. The number of combinations is therefore 10 x 10 = 100.
Each digit can appear in each of the 4 positions. There are 9 digits, therefore there are 9⁴ = 6561 such combinations.
Through the magic of perms and coms the answer is 729
9*9*9 = 729 using the digits 1 to 9 and 2*9 using 10 and another digit. 749 in all.
The answer is 10C4 = 10!/[4!*6!] = 210
120 combinations using each digit once per combination. There are 625 combinations if you can repeat the digits.
It is: 9C7 = 36
66
15
10,000
If there are no restrictions on duplicated numbers or other patterns of numbers then there are 10 ways of selecting the first digit and also 10 ways of selecting the second digit. The number of combinations is therefore 10 x 10 = 100.
Each digit can appear in each of the 4 positions. There are 9 digits, therefore there are 9⁴ = 6561 such combinations.
Through the magic of perms and coms the answer is 729
In other words, how many 4 digit combination locks are there using the digits 0-9 on each wheel. There are 10×10×10×10 = 10⁴ = 10,000 such combinations.
10 Combinations (if order doesn't matter). 3,628,800 Possiblilities (if order matters).
There are 9999 possible combinations starting from 0000 to 9999
6 ways: 931,913,139,193,391,319