The number of combinations possible for taking a specified sub-set of numbers, r, from a set, n, is
C(n,r) = n!/[r!(n-r)!]
In this case, n = 4 and r = 1, so
C(n,r) = 4!/1!3!
C(n,r) = 24/6
C(n,r) = 4
Therefore, there are four possible combinations of the numbers 3, 7, 9 and 4.
==================================
This contributor strongly disagrees, but is leaving the original answer here
so that others can shop and compare.
With their commas, parentheses, and factorials, the formulas are certainly impressive.
Only the conclusion is wrong.
The question specified "4-digit" combinations, so 'n' and 'r' are both 4.
Now, to come down off the pedestal and make it understandable as well as
formally rigorous ...
"Combination" really means different groups of 4 digits that you can select out
of the four you gave us. There's only one of those groups.
If you actually want to know how many different 4-digit numbers you can make
from them, then what you want is called "permutations" of the four digits, and
you can think of it this way, without using 'n', 'r', or parentheses or factorials:
The first digit can be any one of 4. For each of those . . .
The second digit can be any one of the remaining 3. For each of those . . .
The third digit can be either one of the remaining 2.
So the total number of different ways to line them up is (4 x 3 x 2) = 24 different 4-digit numbers.
120 combinations using each digit once per combination. There are 625 combinations if you can repeat the digits.
Any 6 from 51 = 18,009,460 combinations
There are 9 1-digit numbers and 16-2 digit numbers. So a 5 digit combination is obtained as:Five 1-digit numbers and no 2-digit numbers: 126 combinationsThree 1-digit numbers and one 2-digit number: 1344 combinationsOne 1-digit numbers and two 2-digit numbers: 1080 combinationsThat makes a total of 2550 combinations. This scheme does not differentiate between {13, 24, 5} and {1, 2, 3, 4, 5}. Adjusting for that would complicate the calculation considerably and reduce the number of combinations.
Only one.
How many four digit combinations can be made from the number nine? Example, 1+1+2+5=9.
Number of 7 digit combinations out of the 10 one-digit numbers = 120.
There are 38760 combinations.
The answer will depend on how many digits there are in each of the 30 numbers. If the 30 numbers are all 6-digit numbers then the answer is NONE! If the 30 numbers are the first 30 counting numbers then there are 126 combinations of five 1-digit numbers, 1764 combinations of three 1-digit numbers and one 2-digit number, and 1710 combinations of one 1-digit number and two 2-digit numbers. That makes a total of 3600 5-digit combinations.
120 combinations using each digit once per combination. There are 625 combinations if you can repeat the digits.
It is: 9C7 = 36
66
15
6 ways: 931,913,139,193,391,319
10,000
There are 1140 five digit combinations between numbers 1 and 20.
10 Combinations (if order doesn't matter). 3,628,800 Possiblilities (if order matters).
The answer is 10C4 = 10!/[4!*6!] = 210