8 votes 8 votes Let $X$ be a set with $n$ elements. How many subsets of $X$ have odd cardinality? $n$ $2^n$ $2^{n/2}$ $2^{n-1}$ Can not be determined without knowing whether $n$ is odd or even Set Theory & Algebra tifr2019 engineering-mathematics discrete-mathematics set-theory&algebra set-theory + – Arjun asked Dec 18, 2018 • edited May 23, 2019 by go_editor Arjun 3.5k views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
Best answer 14 votes 14 votes $X$ is a set of n elements then there are total $2^n$ subsets out of which $2^{n-1}$ have odd cardinality and $2^{n-1}$ have even cardinality. $\text{D}$ is correct option. Mk Utkarsh answered Dec 18, 2018 • edited Mar 27, 2021 by soujanyareddy13 Mk Utkarsh comment Share Follow See all 0 reply Please log in or register to add a comment.
2 votes 2 votes Quick Proof Consider the binomial expansion of (1-x)^n and put x=0 Yash Khanna answered Dec 22, 2018 • edited Dec 22, 2018 by Shaik Masthan Yash Khanna comment Share Follow See all 0 reply Please log in or register to add a comment.
1 votes 1 votes Following cases must be included if n=0 ans is 0 since cardinality of empty set is zero thus no odd subset if n>0 ans is $2^{n-1}$ number of odd cardinality set Final answer E Tesla! answered Dec 19, 2018 Tesla! comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes Exactly half of the elements of $\mathcal{P}(A)$ are odd-sized. Fix an element $a\in A$ (this is the point where $A\ne\emptyset$ is needed). Then $$S\mapsto S\operatorname{\Delta}\{a\}$$ symmetric difference is a bijection from the set of odd subsets to the set of even subsets. For more on this visit here and here anonymous answered Dec 18, 2018 anonymous comment Share Follow See all 0 reply Please log in or register to add a comment.