From 1,2,3,4,5,6,7, How many combinations without repetition can I get ?
Eg: 1, 2, 3, 4, 5, 6, 7,
1.2,1.3, 1.4, 1.5, 1.6, 1.7 ......
1.2.3, 1.2.4, 1.2.5, 1.2.6 .....
1.2.3.4, 1.2.3.5, 1. 2.3.6...
Until 1.2.3.4.5.6.7
From 1,2,3,4,5,6,7, How many combinations without repetition can I get ?
Eg: 1, 2, 3, 4, 5, 6, 7,
1.2,1.3, 1.4, 1.5, 1.6, 1.7 ......
1.2.3, 1.2.4, 1.2.5, 1.2.6 .....
1.2.3.4, 1.2.3.5, 1. 2.3.6...
Until 1.2.3.4.5.6.7
From 1,2,3,4,5,6,7, How many combinations without repetition can I get ?
Eg: 1, 2, 3, 4, 5, 6, 7,
1.2,1.3, 1.4, 1.5, 1.6, 1.7 ......
1.2.3, 1.2.4, 1.2.5, 1.2.6 .....
1.2.3.4, 1.2.3.5, 1. 2.3.6...
Until 1.2.3.4.5.6.7
I think @Yunzhen is asking about non-empty subsets of \(\displaystyle \{1,2,3,4,5,6,7\}\) s/he is using \(\displaystyle .\) the way \(\displaystyle ,\) is usually used. So \(\displaystyle 2^7-1\) would answer that reading. The total minus one(the empty subset).Are you using the word "combinations" with its technical meaning, or informally? (The technical term implies "without repetition".)
Are you asking about combinations of any number of items (from 1 through 7)? If so, then you might want to call them subsets. There is a simple way to count all subsets of a set. (Then you'll need a small adjustment.)
I think @Yunzhen is asking about non-empty subsets of \(\displaystyle \{1,2,3,4,5,6,7\}\) s/he is using \(\displaystyle .\) the way \(\displaystyle ,\) is usually used. So \(\displaystyle 2^7-1\) would answer that reading. The total minus one(the empty subset).
And I gave the answer, fully in hope that s/he might see the notation and explore further.Yes, I interpreted the question the same way, and subtracting 1 is the "small adjustment" I referred to. I was trying to give only a hint, not the full answer.