how many combinations are in groups of 4 in the numbers 1-9?

Answer 1

I might have misunderstood the question but I would guess its the following:

9999 minus all numbers with zero in them.

Sorry couldn;t be more specific.

Answer 2

It depends on what you mean. "Combinations" properly means individual cases regardless of order (so 1234 is the "same" combination as 4321). The number of combinations of numbers 1 to 9 in groups of four is (9x8x7x6)/(1x2x3x4) = 126. If you meant that numbers can appear more than once (so 1123 is one combination) it is a little more complicatied: there are 126 where all four digits are different, 72 where there are two pairs (9x8), 252 where there is one pair of the same number (9x8x7)/(1x2), 72 where three numbers are the same (9x8), and 9 where are all four numbers are the same. This is a total of 541.
"Permultations" are sometimes meant when somebody refers to "combinations": 1234 and 4321 are different permultations though they are of the same combination. There are 3024 permutations of four different numbers, 432 perms of two pairs, 1310 perms of one pair, 288 perms of three of a kind, and just the 9 perms of all four identical. This gives a total of 4863 different permutations.

About the confusion of perms with combs: note that what we call a "combination lock" is really a "permutation lock" (the order of the numbers is signficant),and what they call "perms" on a pools coupon are actually "combs" (the order of the games selected is insignificant).
By the way somebody else answered "1000" but that embodies TWO mistakes: the first is that this is the calculation for THREE at a time, not four, which would be 10,000. The othe error is that this person included ZERO as a possible number, but you specified only one to nine.

Answer 3

If you are allowed to use the numbers more than once then the answer is 9^4 (9 to the power of 4) ie 6,561. If you're not, then the answer is 9*8*7*6, ie 3024.

Answer 4

If you include 0 and use all the have to have all four numbers then
0000-9999 that would be 10000 different combinations?
Are you trying to pick a bike lock :-)

Answer 5

I'm thinking it's 6561.

Take the number of possible values (9), and raise it to the number of members for the group (4).

This is assuming you are allowing for repeats.

Answer 6

1-9 is 9 numbers. 9c4= 9p4/4! = 9x8x7x6/4x3x2x1 = 3024/24 = about 126

Answer 7

if each number can be used more than once in the combinations 1000