How many 4 digit combinations can be made with numbers 3 7 9 4?

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.

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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.