The answer is 10C4 = 10!/[4!*6!] = 210
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
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.
1,956 different numbers can be made from 6 digits. You can calculate this by using the permutation function in a summation function, like this: Σ6k=1 6Pk = 6P1+6P2+...+6P5+6P6 What this does is calculate how many 1 digit numbers you can make from 6 digits, then how many 2 digit numbers can be made from 6 digits and adds the amounts together, then calculates how many 3 digit numbers can be made and adds that on as well etc.
Not repeating, it is 7*6*5*4 which is 840 ways ---- There are 7 choices for each of four digits, right? 74 = 2401
120 combinations using each digit once per combination. There are 625 combinations if you can repeat the digits.
10 Combinations (if order doesn't matter). 3,628,800 Possiblilities (if order matters).
15
10,000
6 ways: 931,913,139,193,391,319
It is: 9C7 = 36
120 5-digit numbers can be made with the numbers 12345.
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.
The general formula is to square the number of digits it could be (in this case 5) by how many spaces there are. So in this case it would be 53 = 125 possible combinations.
To calculate the number of different 4-digit combinations that can be made using numbers 0 through 9, we use the concept of permutations. Since repetition is allowed, we use the formula for permutations with repetition, which is n^r, where n is the number of options for each digit (10 in this case) and r is the number of digits (4 in this case). Therefore, the number of different 4-digit combinations that can be made using numbers 0 through 9 is 10^4, which equals 10,000 combinations.
Only three: 12, 13 and 23. Remember that the combinations 12 and 21 are the same.