There can be only one combination with four digits. Changing their order gives permutations, NOT combinations.
The following are possible combinations:1 digit: 1, 4, 6.
2 digit: 11, 14, 16, 46
3 digit: 114, 116, 146.
Four outcomes, three combinations.
Four of them.
The question is quite ambiguous, but if I assume you want all possible combinations of the group of the four numbers, then the answer would be 4666, 6466, 6646 and 6664
According to my calculations, with the first four included. It should be 33,396 combinations
If you have 12 possible numbers with multiple combinations then you should start out with making all the possible combinations; you will find theyre 20. Theyre four numbers out of the twleve that can be divisible by three; 3, 6, 9, and 12. There are 7 combinations where the combinations can equal those four numbers. So the odds of getting a sum divisible by three is 7/20.
every number from 0000 to 9999
The number of four-digit combinations is 10,000 .Stick a '3' before each of them, and you have all the possible 5-digit combinations that start with 3.There are 10,000 of them. They run from 30,000 to 39,999 .
combinations
There are 4,500 combinations.
it is i love hunter elam
They are worth 16. There are four different combinations of 5-5-5 and four combinations of J-5. Eight combinations, each worth 2.
18 different combinations. When a coin is tossed twice there are four possible outcomes, (H,H), (H,T), (T,H) and (T,T) considering the order in which they appear (first or second). But if we are talking of combinations of the two individual events, then the order in which they come out is not considered. So for this case the number of combinations is three: (H,H), (H,T) and (T,T). For the case of tossing a die once there are six possible events. The number of different combinations when tossing a coin twice and a die once is: 3x6 = 18 different combinations.