How many combinations can i get 4#s out of 6, such as 1234, 1235,1236, etc and is there a formula to do that?

Question

whats the easiest way to figure the amount of combinations?

Answer 1

u r question isnt clear my friend , u have to give more details on the problem , as u said there is  6 numbers, but u didnt tell whether its already given or not .anyways if the numbers are already given use the formula 6C4 =6!/4!*(6-4)! hope u understand this.

Answer 2

The number of combinations r out of n is

nCr = n!/(r!(n-r)!

in this case n = 6 r = 4 so we have 6!/4!2! = 6*5/2 or 15

Answer 3

hi james...

if 1234 and 2134 are two of the solutions then...

i think what you mean is permutation not combination. YES they HAVE the formulae.

the formula is nPr = n! / (n-r)!

in this case you have 6 options, so n= 6,

then you have to choose 4 from 6, so r =4

to get how many possible arrangements = 6P4 = 6! / (6-4)! = 3x4x5x6=360 arrangements.

see ya james...

Answer 4

You can choose the first number 6 ways, the second 5 ways, the third 4 ways and the fourth 3 ways

The total number of different choices is thus 6x 5x 4x 3 =360

This is 6!/2!  

Answer 5

I don't see anything that says they have to be in numerical order. If not, then do it this way.

1st number; 6 choices

2nd number 5 choices, 6 x 5 = 30

3rd, 4 choices left, 30 x 4 = 120

last, 3 choices left, 120 x 3 = 360 total combinations.  

Answer 6

I think you're looking for the number "6 choose 4", which is also called a "binomial coefficient".

On a calculator, it might look like C(n,k) or nCk, though there are variations.

6C4 = 6! / (4! 2!), where:

6! = 6*5*4*3*2*1 = 720

4! = 4*3*2*1 =24

2! = 2*1 = 2

The 2 in 2! is there because 2+4=6.

To make a long story short, there are 15 combinations.  That's not a whole lot, so I'll list them:

1234, 1235, 1236, 1245, 1246,

1256, 1345, 1346, 1356, 1456,

2345, 2346, 2356, 2456, 3456