There are 1140 5 digit combinations from 1 to 20. 20 combination 3 computes that.
There are 23C3 = 23!/(20!*3!) = 23*22*21/(3*2*1) = 1,771 combinations.
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.
There are 10 choices for the first number, 9 for the second and 8 for the third. 10*9*8=720 possible combinations.
Well, honey, there are 20 numbers to choose from for the first digit, 19 for the second, and 18 for the third. So, multiply those together and you get a total of 6,840 possible 3-number combinations. Math can be a real party pooper, but hey, that's the answer for ya!
There are 6C3 = 20 such combinations.
4
There are 1140 5 digit combinations from 1 to 20. 20 combination 3 computes that.
There are 23C3 = 23!/(20!*3!) = 23*22*21/(3*2*1) = 1,771 combinations.
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.
The number of combinations is 20C5 = 20!/(15!*5!) = 20*19*18*17*16/(5*4*3*2*1) = 15,504
Combinations of 2: 20*19/(1*2) = 190 Combinations of 3: 20*19*18/(1*2*3) = 1140 Combinations of 4: 20*19*18*17/(1*2*3*4) = 4845 So 6175 in all.
There are 10 choices for the first number, 9 for the second and 8 for the third. 10*9*8=720 possible combinations.
Well, honey, there are 20 numbers to choose from for the first digit, 19 for the second, and 18 for the third. So, multiply those together and you get a total of 6,840 possible 3-number combinations. Math can be a real party pooper, but hey, that's the answer for ya!
63
2 x 3 x 3 = 18
If their sequence matters . . . 120 If it doesn't . . . 20