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Apologies if my terminology is incorrect, as I don't have a mathematics background. I'm looking for a formula to work out the number of combinations in a sequence - but where any number of the items can be used. The order does not matter.

For example, for:

A,B,C

The combinations I'm looking for are:

A
B
C
AB
AC
BC
ABC

Combinations seems to be the closest thing to what I'm looking for (nCr), but I need it where r can equal any number.

Neither combinations nor permutations seem to provide me with what I'm looking for, and I've run out of ideas trying to find what the correct mathematical terminology is, and what the formula would be for working this out.

Thank you!

Chris Jenkins
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    Not sure this is clear. Are $ABC$ and $CBA$ the same or different? If the same, then you just want the number of non-empty subsets (so $2^n-1$). Or did you mean something else? ā€“ lulu Jul 25 '16 at 13:54
  • Thanks, ABC and CBA are the same, the order is irrelevant. 2^nāˆ’1 does indeed seem to be what I was looking for! Thank you! ā€“ Chris Jenkins Jul 25 '16 at 14:02

2 Answers2

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If we choose k elements from a n-set in $\binom{n}{k}$ ways we get all k-combinations without repetition, from your example for $n=3$ and $k=1,2,3$ we have $$\binom{3}{1}+ \binom{3}{2}+ \binom{3}{3}=3+3+1=7$$ in general case

$$\sum_{k=1}^{n}\binom{n}{k}=2^n-1$$

Adi Dani
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While trying to take 2 objects out of 3 objects(in way that you do not care about the sequence),we use combination.

$\binom{3}{2} = \frac{3!}{2!(3-2)!}=3$....(i)

Why do we use combination?

Example:

You have 3 footballs.You need 2.And there are red,blue and green colored football.

First,You look at the footballs.And I ask you a question:"In how many ways you can take a football from here?"

ANSWER:3(You can take red or blue or green one)

Now you need another ball. But now you have only 2 options.

The no. of way you can select the first ball=$3$

The no. of way you can select the second ball=$2$

The no. of way you can select the two ball$=3*2=6$

BUT, since we do not care about the sequence of selecting balls,we have a repetition of the same results.

(red,blue) is same as (blue,red) [Taking red ball first and blue ball second is same as taking the balls in reverse sequence]

Therefore,The no. of way you can select the two ball$=(3*2)/2=6/2=3$(which is same as (i))

Now use the same method to take 1 object out of 3 and 3 objects out of 3.

Notice that the events of taking '1 object out of 3' and '2 objects out of 3' and '3 objects out of 3' ARE NOT SIMULTANEOUS,they are independent events.

So,we sum the possibilities to find the no. of taking '1 object out of 3' or '2 objects out of 3' or '3 objects out of 3:

$\binom{3}{3}+ \binom{3}{2}+\binom{3}{1}= \frac{3!}{3!(3-3)!}+\frac{3!}{2!(3-2)!}+\frac{3!}{1!(3-1)!}=3+3+1=7$

Look,we have seven combination.These seven combinations are

A

B

C

AB

AC

BC

ABC