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.
This questions can be rewritten as 52 choose 6 or 52C6. This is the same as (52!)/(6!(52-6)!) (52!)(6!46!) (52*51*50*49*48*47)/(6*5*4*3*2*1) 14658134400/720 20358520 There are 20,358,520 combinations of 6 numbers in 52 numbers. This treats 1,2,3,4,5,6 and 6,5,4,3,2,1 as the same combination since they are the same set of numbers.
Since that's a fairly small set, you should be able to check all combinations (for 2 numbers, there are only 4 possible multiplications), and see whether the result is in the set.
The mean is the is the total of the numbers and then dividing by how many numbers.
The mean of a set of numbers is found by adding them together and dividing by how many numbers were added.
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.
41
By making a number tree that could have as many as 1,000,000 combos.
Could you generate a complete set of 6 number combinations from 45 numbers ?
86,450
It gives you the amounts of combinations of numbers from a set of numbers. For example if you wanted to find how many combinations of 2 numbers can you get from 6 numbers you would do this: =COMBIN(6,2) It will give the result 15, because there are 15 combinations as follows, where we are getting combinations of 2 numbers from the numbers 1 to 6: 1,2 1,3 1,4 1,5 1,6 2,3 2,4 2,5 2,6 3,4 3,5 3,6 4,5 4,6 5,6
To calculate the number of 7-number combinations from 8 numbers, you can use the combination formula, which is nCr = n! / r!(n-r)!. In this case, n = 8 (total numbers) and r = 7 (numbers chosen). Plugging these values into the formula, you get 8C7 = 8! / 7!(8-7)! = 8 ways. Therefore, there are 8 different combinations of 7 numbers that can be chosen from a set of 8 numbers.
9999 + 1 * * * * * That is so wrong! There are 10*9*8*7/(4*3*2*1) = 210 combinations of 4 numbers from the set of ten 1-digit numbers, 0 1, 2, ... 9 When considering combinations, 1234 is the same as 2314 or 4123 etc. Actually... 4 numbers to be arranged with ten different possibilities each will give a total of 10^4 combinations, or 10,000.
To answer this question you need to start with the "... combinations you wrote for 12, 18 and 21". Notice that the numbers in the second set are each 10 times the numbers in the first set. Combinations for numbers in the second set can be found by adding "*10" or "*5*2" to the combinations for the first set. You could also say: 12 = 2*2*3; 120 = 2*2*3*5*2 = 4*15*2, and many other variations on this theme.
If the numbers are all different, then 25C5 = 25*24*23*22*21/(5*4*3*2*1) = 53,130.
The possible combinations of a set of 60 different numbers would be 60! or 60 factorial. This is a very number at 8.3209871 x 10 ^ 81, or a figure with 81 zeroes behind it. The official short scale name would be 8.3209871 "Sesvigintillion".
This questions can be rewritten as 52 choose 6 or 52C6. This is the same as (52!)/(6!(52-6)!) (52!)(6!46!) (52*51*50*49*48*47)/(6*5*4*3*2*1) 14658134400/720 20358520 There are 20,358,520 combinations of 6 numbers in 52 numbers. This treats 1,2,3,4,5,6 and 6,5,4,3,2,1 as the same combination since they are the same set of numbers.