10 * * * * * That is just plain wrong! It depends on how many numbers in each combination but there are 1 combination of 4 numbers out of 4, 4 combinations of 3 numbers out of 4, 6 combinations of 2 numbers out of 4, 4 combinations of 1 number out of 4. A grand total of 15 (= 24-1) combinations.
7
To calculate the number of 4-number combinations possible with 16 numbers, you would use the formula for combinations, which is nCr = n! / r!(n-r)!. In this case, n = 16 (the total number of numbers) and r = 4 (the number of numbers in each combination). Plugging these values into the formula, you would calculate 16C4 = 16! / 4!(16-4)! = 1820. Therefore, there are 1820 possible 4-number combinations with 16 numbers.
6C4 = 6*5/(2*1) = 15
It depends on whether the order of the numbers is important or not. For example, if 123456 is seen as a different code to 213456 then there are many more possible solutions.If the order is unimportant, the number of possible combinations is equal tobinomial coefficient(15,6) = 5005If the order is important, then the number of possible permutations is equal to15! = ~1.3x1012
There are 15 combinations.
it is hard to say there are lot of combinations belive or not * * * * * If the previous answerer thinks 15 is a lot then true. There are 1 combination of 4 numbers out of 4, 4 combinations of 3 numbers out of 4, 6 combinations of 2 numbers out of 4, 4 combinations of 1 number out of 4. A grand total of 15 (= 24-1) combinations. Not so hard to say!
14 * * * * * Wrong! There are 15. 4 combinations of 1 number, 6 combinations of 2 number, 4 combinations of 3 numbers, and 1 combination of 4 numbers.
Assuming that the six numbers are different, the answer is 15.
There are infinite combinations that can make 3879
The number of combinations is 20C5 = 20!/(15!*5!) = 20*19*18*17*16/(5*4*3*2*1) = 15,504
The number of 4 different book combinations you can choose from 6 books is;6C4 =6!/[4!(6-4)!] =15 combinations of 4 different books.
all numbers have to be different 720 360 (6*5*4*3*2*1), * * * * * Wrong. Combinations do not have to be different. Those are permutations. There are only 6C4 = 6*5/(2*1) = 15 combinations.
10 * * * * * That is just plain wrong! It depends on how many numbers in each combination but there are 1 combination of 4 numbers out of 4, 4 combinations of 3 numbers out of 4, 6 combinations of 2 numbers out of 4, 4 combinations of 1 number out of 4. A grand total of 15 (= 24-1) combinations.
There are 18C4 = 18!/[18-4)!4!] = 18*17*16*15/(4*3*2*1) = 3060 combinations.
15 choose 3 = 15!/(3!*(15-3)!) = 15!/(3!*12!) = (15*14*13)/(3*2) = 5*7*13 = 455
32,432,400 non-unique combinations (i.e., combinations that may be duplicated). To determine how many combinations there are, figure out how many choices can be had for each pick and multiply them together. 15 numbers and one can pick 7. First pick is 1 of 15. Second pick is 1 of the remaining 14. ... Seventh pick is 1 of the remaining 9. So: 15 x 14 x 13 x 12 x 11 x 10 x 9 = 32,432,400.