Assuming that the eight numbers are all different and that none of them are zero, the number of combinations is 8C3 = 8!/(3!*5!) where n! = 1*2*3*...*n the number is 8*7*6 / (3*2*1) = 56
There are only four combinations but there are 8 permutations.
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.
There are 10 choices for the first number, 9 for the second and 8 for the third. 10*9*8=720 possible combinations.
Six * * * * * No, that is the number of PERMUTATIONS (not combinations). With 3 numbers, the number of combinations, including the null combination, is 23 = 8. With the three numbers 1,2 and 3, these would be {None of them}, {1), {2), {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3}.
Assuming that the eight numbers are all different and that none of them are zero, the number of combinations is 8C3 = 8!/(3!*5!) where n! = 1*2*3*...*n the number is 8*7*6 / (3*2*1) = 56
You can make 5 combinations of 1 number, 10 combinations of 2 numbers, 10 combinations of 3 numbers, 5 combinations of 4 numbers, and 1 combinations of 5 number. 31 in all.
The number of combinations is 10C4 = 10*9*8*7/(4*3*2*1) = 210
There are only four combinations but there are 8 permutations.
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.
There are 10 choices for the first number, 9 for the second and 8 for the third. 10*9*8=720 possible combinations.
Six * * * * * No, that is the number of PERMUTATIONS (not combinations). With 3 numbers, the number of combinations, including the null combination, is 23 = 8. With the three numbers 1,2 and 3, these would be {None of them}, {1), {2), {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3}.
There are 10 combinations of 1 number,10*9/(2*1) = 45 combinations of 2 numbers,10*9*8/(3*2*1) = 120 combinations of 3 numbers,10*9*8*7/(4*3*2*1) = 210 combinations of 4 numbers,10*9*8*7*6/(5*4*3*2*1) = 252 combinations of 5 numbers,210 combinations of 6 numbers,120 combinations of 7 numbers,45 combinations of 8 numbers,10 combinations of 9 numbers, and1 combination of 10 numbers.All in all, 210 - 1 = 1023 combinations. I have neither the time nor inclination to list them all.There are 10 combinations of 1 number,10*9/(2*1) = 45 combinations of 2 numbers,10*9*8/(3*2*1) = 120 combinations of 3 numbers,10*9*8*7/(4*3*2*1) = 210 combinations of 4 numbers,10*9*8*7*6/(5*4*3*2*1) = 252 combinations of 5 numbers,210 combinations of 6 numbers,120 combinations of 7 numbers,45 combinations of 8 numbers,10 combinations of 9 numbers, and1 combination of 10 numbers.All in all, 210 - 1 = 1023 combinations. I have neither the time nor inclination to list them all.There are 10 combinations of 1 number,10*9/(2*1) = 45 combinations of 2 numbers,10*9*8/(3*2*1) = 120 combinations of 3 numbers,10*9*8*7/(4*3*2*1) = 210 combinations of 4 numbers,10*9*8*7*6/(5*4*3*2*1) = 252 combinations of 5 numbers,210 combinations of 6 numbers,120 combinations of 7 numbers,45 combinations of 8 numbers,10 combinations of 9 numbers, and1 combination of 10 numbers.All in all, 210 - 1 = 1023 combinations. I have neither the time nor inclination to list them all.There are 10 combinations of 1 number,10*9/(2*1) = 45 combinations of 2 numbers,10*9*8/(3*2*1) = 120 combinations of 3 numbers,10*9*8*7/(4*3*2*1) = 210 combinations of 4 numbers,10*9*8*7*6/(5*4*3*2*1) = 252 combinations of 5 numbers,210 combinations of 6 numbers,120 combinations of 7 numbers,45 combinations of 8 numbers,10 combinations of 9 numbers, and1 combination of 10 numbers.All in all, 210 - 1 = 1023 combinations. I have neither the time nor inclination to list them all.
The answer is the number of combinations of 7 vertices from 10.That is 10C7 = 10!/(7!*3!) = 10*9*8/(3*2*1) = 120.And, there is no such word as verticles.The answer is the number of combinations of 7 vertices from 10.That is 10C7 = 10!/(7!*3!) = 10*9*8/(3*2*1) = 120.And, there is no such word as verticles.The answer is the number of combinations of 7 vertices from 10.That is 10C7 = 10!/(7!*3!) = 10*9*8/(3*2*1) = 120.And, there is no such word as verticles.The answer is the number of combinations of 7 vertices from 10.That is 10C7 = 10!/(7!*3!) = 10*9*8/(3*2*1) = 120.And, there is no such word as verticles.
Since there are 3 numbers to choose from, there will be 2^3 = 8 combinations. This includes the null combinations which you may not wish to include in the count.
There are 10C3 = 10*9*8/(3*2*1) = 120 combinations.
Infinitely many. Only 6 with both, the number of 8s and the number of 3s being positive.