Question

A sel contains eleven elements.

Answer 1

Answers:

• a) 2048
• b) 2047

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Explanation:

If a set has n elements inside it, then it has 2^n different subsets.

Consider a set like {a,b,c}. It has n = 3 elements.

It has 2^n = 2^3 = 8 subsets

Those 8 subsets are...

1. {a,b,c} ..... any set is a subset of itself
2. {a,b}
3. {a,c}
4. {b,c}
5. {a}
6. {b}
7. {c}
8. { } .... the empty set

As you can see, each subset consists of items that are selected from the original set {a,b,c}. We can't have any repeat letters.

In that list, items 2 through 4 represent subsets with exactly two things inside it. Items 5 through 7 are known as singletons as they only have one item inside each set. The empty set can be written with the symbol $\varnothing$ or you could have a pair of braces with nothing inside them. The empty set is a subset of any set.

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Note how I included {a,b,c} as a subset. Any set is a subset of itself.

A proper subset will ignore the original set and only look at smaller subsets. If we say B is a proper subset of A, then set B will have fewer items compared to set A. Without the "proper" in there, it's possible that A = B.

So all we've done really is kick out one set which drops 2^n to (2^n)-1 when counting the number of proper subsets. I'm using parenthesis to indicate the "-1" is not part of the exponent. If you wrote this on your paper, then you would likely write $2^{n} - 1$

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For this problem, n = 11

This means there are 2^n = 2^11 = 2048 subsets and (2^n)-1 = 2048-1 = 2047 proper subsets.