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.
There are 13,244 of them.
There are 7C4 = 7!/(4!*3!) = 7*6*5/(3*2*1) = 35 combinations.
There are 1285C4 = 1285*1284*1283*1282/(4*3*2*1) = 113076300485 combinations.
There are 11,238,513 of them.
I am assuming you mean 3-number combinations rather than 3 digit combinations. Otherwise you have to treat 21 as a 2-digit number and equate it to 1-and-2. There are 21C3 combinations = 21*20*19/(3*2*1) = 7980 combinations.
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.
There are 13,244 of them.
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}.
4*3*2*1 = 24 different combinations.
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.
To calculate the number of 3-number combinations from the numbers 1-20, we can use the formula for combinations, which is nCr = n! / r!(n-r)!. In this case, n = 20 and r = 3. Plugging these values into the formula, we get 20C3 = 20! / 3!(20-3)! = 1140. Therefore, there are 1140 possible 3-number combinations from the numbers 1-20.
There are 120 permutations and 5 combinations.
The number of combinations is 10C4 = 10*9*8*7/(4*3*2*1) = 210
How many four digit combinations can be made from the number nine? Example, 1+1+2+5=9.
39*38*37/(3*2*1) = 9139
1