two
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 there is no repetition, you want the number of groups of 4 numbers, and the order doesn't matter: this is calculated as 33 over 4; which is the same as (33 x 32 x 31 x 30) / (1 x 2 x 3 x 4)
4!=4x3x2x1=24
1
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.
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 33C6 = 33*32*31*30*29*28/(6*5*4*3*2*1) = 1,107,568 combinations.
31C5 which is 31!/[5!*(31-5)!] = 31*30*29*28*27/(5*4*3*2*1) = 169911
16
You would get 4!/2! = 12 combinations.
8C4 = 70
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.
If the numbers are allowed to repeat, then there are six to the fourth power possible combinations, or 1296. If they are not allowed to repeat then there are only 360 combinations.
There are 15 combinations.
4 of them. In a combination the order of the numbers does not matter.