Like 1235 or 8067?
If you allow leading zeros, then 10 000.
If you disallow leading zeros, then 9 000.
* * * * *
Those are the numbers of PERMUTATIONS, not COMBINATIONS.
In combinations the order does not matter, so 1234 = 4312 = 2314 etc.
The number of combinations is (10*9*8*7)/(4*3*2*1) = 210.
The question of leading 0s does not arise because there is no order to the set of numbers in the combination so there is no "leading".
10
How many four digit combinations can be made from the number nine? Example, 1+1+2+5=9.
To calculate the number of 3-digit combinations that can be made from the numbers 1-9, we can use the formula for permutations. Since repetition is allowed, we use the formula for permutations with repetition, which is n^r, where n is the total number of options (10 in this case) and r is the number of digits in each combination (3 in this case). Therefore, the total number of 3-digit combinations that can be made from the numbers 1-9 is 10^3 = 1000.
All the numbers from 0 to 999 so 1000 numbers However if you mean only using each number once the answer is about 720 * * * * * No. The above answer refers to the number of permutations. Permutations are NOT the same as combinations as anyone who has studies any probability theory can tell you. The number of combinations is 9C3 = 9*8*7/3*2*1 = 84
6
45
The number of combinations of six numbers that can be made from seven numbers will depend on if you can repeat numbers. In all there are over 2,000 different numbers that can be made.
10
How many four digit combinations can be made from the number nine? Example, 1+1+2+5=9.
2 to the 7th power = 128 * * * * * No. That is the total number of combinations, consisting of any number of elements. The number of 2 number combinations is 7*6/2 = 21
9!/5! or 3024
8C6 = 28
None. You do not have enough numbers to make even one combination.
Using the formula n!/r!(n-r)! where n is the number of possible numbers and r is the number of numbers chosen, there are 13983816 combinations of six numbers between 1 and 49 inclusive.
20
If repetition is allowed and order is important, then you have essentially a base-4 number system, with the numbers ranging from 00004 to 33334. The quantity of permutations in this example is 44 = 256. If repetition is not allowed, but order is important, then it is 4! = 24. * * * * * The above answer is perfectly correct. But, as stated in the answer, for permutations. However, according to the mathematical definition of combinations (as opposed to permutations), the order is irrelevant to combinations. 1234 is the same as 1423 or 4213 etc. Consequently, there can be only one 4-number combination from 4 numbers
There is 1 combinations with 7 numbers in them,7 combinations with 1 or 6 numbers in them,21 combinations with 2 or 5 numbers in them,35 combinations with 3 or 4 numbers in them.Notionally, there is also one combination with no numbers in it.In all then, there are 128 ( = 27) combinations.There is 1 combinations with 7 numbers in them,7 combinations with 1 or 6 numbers in them,21 combinations with 2 or 5 numbers in them,35 combinations with 3 or 4 numbers in them.Notionally, there is also one combination with no numbers in it.In all then, there are 128 ( = 27) combinations.There is 1 combinations with 7 numbers in them,7 combinations with 1 or 6 numbers in them,21 combinations with 2 or 5 numbers in them,35 combinations with 3 or 4 numbers in them.Notionally, there is also one combination with no numbers in it.In all then, there are 128 ( = 27) combinations.There is 1 combinations with 7 numbers in them,7 combinations with 1 or 6 numbers in them,21 combinations with 2 or 5 numbers in them,35 combinations with 3 or 4 numbers in them.Notionally, there is also one combination with no numbers in it.In all then, there are 128 ( = 27) combinations.