how many combinations are there for 3 lots of 1-10 numbers?

Question

3 lots of numbers from 1,2,3,...8,9,0
how many 3 digits combinations are there
i.e. take 1 out of the first lot and take 4 out of the second lot and 6 from the 3rd, the combinataion becomes 146...how many different combinations are there?

Answer 1

10^3 = 1000 combinations

Answer 2

Assumming you can only use each number once then the answer is 10! / 7! which in longhand is

10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 divided by 7 * 6 * 5 * 4 * 3 * 2 * 1.

With the numbers 1 thru 7 both above and below the line, they cancel each other out making the answer 10 * 9 * 8 = 720.


The equation to find any combination of numbers is X! / (X - n)! where X is the number of options and n is the number of choices. For example if you wanted to find the number of combination any 4 letters of the alphabet could make, the answer would be 26! / (26-4)! which is the same as 26 * 25 * 24 * 23 which = 358800

Answer 3

The number of combinations of three things taken from ten (assuming the ten are all different) is 120. Others' answers of 720 are the number of permutations. Perhaps, of course, this is what the questioner meant, but that is not stated.

And I interpreted the question differently anyway. I took it as how many combinations are there if you take one number from ten, then one number from another ten, then one more from another ten. This makes it possible for there to be duplicate digits.

In that case there are 120 combinations of three different digits, 90 combinations where one digit in doubled, and ten where all three are the same. This totals 220, so I think that is the answer to the original question...

... if I have interpreted the question correctly.

Answer 4

10^3 = 1000.

Can the same number be used again? If cannot, then it becomes 10 x 9 x 8 = 720

Answer 5

to find any 3 from10;
10C3=10!/(10!-3!)*3!
=8*9*10/1*2*3=120
if the three are in a particular
order{permutation},it is
10!/7!=8*9*10
=720

summary:-
to find the number of
combinations of n different
things taken r together;
if we take any one of the
combinations,then the r
things which it contains
can be arranged in r!
different orders
if this is done to all the
combinations,we shall have
arranged all the groups of r
things,that is, we shall have
permuted them
therefore,
nCr*r!=nPr=n!/(n-r)!
hence,
nCr=n!/[(n-r)!*r!]
{in this problem,
n=10,r=3}

i hope that this helps