Well, assuming you mean that there are 37 possible numbers you'd follow this method: c = 37 x (37-1) x(37-2) x (37-3) x ... (37-(n-1)) where c is the number of combinations and n is the number of numbers drawn. For example: if n = 6 then: c = 37 x 36 x 35 x 34 x 33 x 32 = 1673844480 combinations. There's normal a button on your calculator that'll do this for you.
There is only one five-number combination of 5 distinct numbers.
Just one. * * * * * Depends on how many numbers are on each ring. If there are x numbers, then the total number of combinations (actually they are permutations) is x*x*x or x3.
If you only include unique answers you end up with the following 6 combinations:0088, 0808, 8008, 8800, 8080, and 0880
This is an ambiguous question. Are you asking for the number of combinations, or to actually list them? Since we haven't been given the 8 numbers, I assume the former. Are you asking for the number of combinations of 4 different numbers selected from the given 8, or are duplicates allowed? The latter is kind of complicated. The number of permutations of 4 numbers selected from 8 given numbers (or any 8 things) is very easy: since each of the 4 positions in the sequence can be any of the 8, you get 8 raised to the 4th power, which is 4096. But to get the number of combinations, you have to eliminate the duplications, and the number of duplications for a permutation depends on the number of repetitions within the permutation itself. For example, the permutation (1, 2, 1, 2) has 2 pairs; you need to count the number of combinations of the 4 positions where the 1s can go, and the other 2 positions would have 2s. That number turns out to be 6, but if there is only 1 pair, as in (1, 1, 2, 3) you have 6 pairs of positions to put the 1s in, and 2 ways of ordering the 2 and the 3, so you get 12. I could go on and on. If you don't allow duplications, then you have 8 x 7 x 6 x 5 possible permutations, which is 1,680. Each of these can be ordered 4 x 3 x 2 x 1 ways, which is 24. Dividing 1,680 by 24 gives 70 combinations
There are 21000 - 1 combinations. Remember that in a combination the order of the numbers does not matter. That is relevant only for permutations which are not the same thing. For each combination, each one of the 1000 numbers can be in or out. So for each combination there are 1000 choices of in or out. Except that if all are out, you get a null combination - that is, a combination with no numbers. For that particular case you subtract the 1 from the total. You will have 1 combination consisting of 1000 numbers, 499500 of 2 numbers and so on. The number of combinations will be the nth row of Pascal's triangle.
It isn't possible. Quite simply, if there were a way to do it, such a method would spread very quickly and that would ruin the lottery; there would be no more lottery.
if i got everything right, its 1000
Assuming that repeated numbers are allowed, the number of possible combinations is given by 40 * 40 * 40 = 64000.If repeated numbers are not allowed, the number of possible combinations is given by 40 * 39 * 38 = 59280.
If the numbers can be repeated and the numbers are 0-9 then there are 1000 different combinations.
If the order of the numbers are important, then this is a simple combination problem. There are 10 possible numbers to choose from for the first number. Then there are 9 options for the second number. Then there are 8 options for the third, and so on. Thus, the number of possible combinations can be calculated as 10x9x8x7x6x5. This comes out at 151,200 possible combinations.
There are infinitely many numbers and so infinitely many possible combinations.
Out of the 42 available numbers, you simply select the 6 numbers that aredrawn and announced as the winning combination. Note that you must makeyour selection before the winning combination is drawn.There are only 5,245,786 different possible combinations.
4 of them. In a combination the order of the numbers does not matter.
Just 4: 123, 124, 134 and 234. The order of the numbers does not matter with combinations. If it does, then they are permutations, not combinations.
At this point it is not possible to tell the two number VN lottery going out today.
The number of combinations of 50 things taken 5 at a time is (50! - 45!) / 5! or 2,118,760, so the probability of winning the lottery on 1 ticket by picking 5 numbers out of 50 numbers is 1 in 2,118,760, or 0.00000047197. More formally, the number of combinations of N things taken P at a time is (N! - (N-P)!) / P!
figure it out then tell me :)