combinations of 7 numbers using 10 numbers?

Answer 1

The combination of n things taken r at a time:
nCr = n!/(r!(n-4)!)

So your problem:
10 C 7 = 10!(7!·3!)
= (10·9·8·7·6·5·4·3·2·1)/(3·2·1·7·6·5·4·3·2·1)
=(10·9·8)/(3·2·1)
=720/6 = 120

Answer 2

How many ways to pick 7 numbers out of a collection of 10?

Also written C(7,10)

the first number can be any one of 10 (10 ways)
Once this number is picked, the second one can be any of the remaining 9, etc.

Total number of ways =
10*9*8*7*6*5*4 = 604,800

In this answer, these two combinations are NOT the same;
8,3,1,4,9,2,6
3,8,1,4,9,2,6

If the order does not matter (i.e., if you consider 8,3 the same as 3,8 as the first two choices), then for any combination of 7 numbers, there are FACT(7) ways to arrange them:
7*6*5*4*3*2*1 = 5040 ways.

Therefore, there are 604,800 / 5,040 = 120 ways to pick 7 numbers out of 10.

For big numbers, use the formula:
How many ways to pick x numbers out of N:

FACT(N) / [ FACT(x)*FACT(N-x)]

In this problem:

FACT(10) / [FACT(7)*FACT(3)]

3,628,800 / [5,040*6]
3,628,800 / 30,240 = 120

Answer 3

I think you need to make your question a little bit more clear...

Do you mean like:

12, 3, 4, 5, 6, 78, 90

Or something else?

Answer 4

10C7 = 10C3 = 10*9*8/3! = 120