Combinations of things: 7 items; numbers of combos of 2, 3, etc w/o double-ups

[NekoPieChan]

So I have 7 items and I would like to know how many combinations of them I can get without double ups or anything.

Say I have the items as - A, B, C, D, E, F, G,

I want to know the amount combinations of 2 items, 3 items, 4 items, 5 items, and 6 items, without double ups (like A-B-C doubled as C-B-A, A-C-B, C-A-B, B-A-C, B-C-A).

I already calculated 2 items to being 21 combinations and 3 items to being 37 combinations but I may have messed up somewhere.

Any help would be appreciated

 

[pka]

So I have 7 items and I would like to know how many combinations of them I can get without double ups or anything.
Say I have the items as - A, B, C, D, E, F, G,
I want to know the amount combinations of 2 items, 3 items, 4 items, 5 items, and 6 items, without double ups (like A-B-C doubled as C-B-A, A-C-B, C-A-B, B-A-C, B-C-A).

I assume that you do not want to consider the empty set and all singleton sets. There are eight of those.
There are a total of \(\displaystyle 2^7 \) total subsets. Thus how many contain two or more?

If you use combinations the answer is \(\displaystyle \sum\limits_{n = 2}^7 \dbinom{7}{n} \)

 

[Harry_the_cat]

So I have 7 items and I would like to know how many combinations of them I can get without double ups or anything.

Say I have the items as - A, B, C, D, E, F, G,

I want to know the amount combinations of 2 items, 3 items, 4 items, 5 items, and 6 items, without double ups (like A-B-C doubled as C-B-A, A-C-B, C-A-B, B-A-C, B-C-A).

I already calculated 2 items to being 21 combinations and 3 items to being 37 combinations but I may have messed up somewhere.

Any help would be appreciated

Ok, choosing 2 things. If you have 7 things, there are 7 ways of choosing the first thing and then 6 ways of choosing the second. 7*6 =42
HOWEVER, this counts eg AB and BA as two different ways. So you need to divide by 2 ie \(\displaystyle \frac{7*6}{2*1}\) = 21, like you said.

Now consider choosing 3 things. There are 7 ways of choosing the first, 6 ways of choosing the second and 5 ways of choosing the third. 7*6*5=210.
BUT, this counts ABC, ACB, BAC, BCA, CAB, CBA as different ways. So you need to divide by 6 (that's the number of ways you can arrange ABC...ie 3*2*1)
So you get \(\displaystyle \frac{7*6*5}{3*2*1}\) = 35 ways. (You got 37 so you must have doubled up somewhere.

You should be seeing a pattern here. What do you think will be the number of ways of choosing 4 items etc?

BTW, this is what the nCr button on your calculator does - gives you how many ways you can pick r things out of 7 (where order is not important).