To calculate the number of combinations that can be made with the numbers 3, 6, and 9, we can use the formula for combinations without repetition, which is nCr = n! / (r!(n-r)!). In this case, we have 3 numbers and we are choosing all of them, so n=3 and r=3. Plugging these values into the formula, we get 3C3 = 3! / (3!(3-3)!) = 6 / (6*1) = 1 combination. Therefore, there is only 1 combination that can be made with the numbers 3, 6, and 9.
6 of them.
6
6
There are 6!/3! = 120 different 3-digit numbers that can be made from these 6 digits.
10C6 = 10*9*8*7/(4*3*2*1) = 210 combinations.
6 different combinations can be made with 3 items
6 of them.
6
6
6! = 6*5*4*3*2*1 = 720
There are 1,120,529,256 combinations.
There are 6!/3! = 120 different 3-digit numbers that can be made from these 6 digits.
6 ways: 931,913,139,193,391,319
There are 6C3 = 20 such combinations.
3! or 6 combinations can be made from three distinct numbers. For this example they are: 345, 354, 534, 543, 435, 453.
7C3 = 7*6*5/(3*2*1) = 35 combinations.7C3 = 7*6*5/(3*2*1) = 35 combinations.7C3 = 7*6*5/(3*2*1) = 35 combinations.7C3 = 7*6*5/(3*2*1) = 35 combinations.
6