56 combinations. :)
Chat with our AI personalities
There are 167960 combinations.
10 possible numbers on each wheel equals 10x10x10 or 1000 combinations possible.
To calculate the combinations for the numbers 2, 6, and 8, we need to use the formula for combinations, which is nCr = n! / r!(n-r)!. In this case, we have 3 numbers (n=3) and we want to choose 2 of them (r=2). So, the combinations would be 3C2 = 3! / 2!(3-2)! = 3. Therefore, the combinations for 2, 6, and 8 are (2, 6), (2, 8), and (6, 8).
Just one. unless you count 123456 different from 132456 then there are 46656 * * * * * But you cannot count 123456 as different from 132456 since it is NOT a different combination. And the question was about combinations.
Number of 7 digit combinations out of the 10 one-digit numbers = 120.
10,000
If the numbers can be repeated and the numbers are 0-9 then there are 1000 different combinations.
If you can repeat the numbers within the combination there are 10,000 different combinations. If you cannot repeat the numbers within the combination, there are 5040 different combinations.
There are 167960 combinations.
The four possible combinations are:A = (+, +)B = (+, -)C = (-, +) andD = (-, -)In A and D, the two numbers have the same signs and the multiplication gives a positive answer.In B and C, the two numbers have different signs and the multiplication gives a negative answer.
Just 1.
The rearrangement of 5 figure numbers will be 5x4x3x2x1 which is 120 combinations, when you don't repeat a number.
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.
Assuming 9 numbers chosen from 56, with no repetition allowed, there are 7575968400 possible combinations.
If the numbers are allowed to repeat, then there are six to the fourth power possible combinations, or 1296. If they are not allowed to repeat then there are only 360 combinations.
there are 13,983,816 combinations.
If the order of the numbers are important, then this is a simple combination problem. There are 10 possible numbers to choose from for the first number. Then there are 9 options for the second number. Then there are 8 options for the third, and so on. Thus, the number of possible combinations can be calculated as 10x9x8x7x6x5. This comes out at 151,200 possible combinations.