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from set $S=${$1,2,3.....n$} in how many ways can we select two subsets $A$, $B$ such that $B$ is non empty and is contained in $A$ which is contained in $S$ And the number of cases in which the containment is proper. I dont need the answer just now but just hints I want to try the question myself. The basic issue is that i am not able to understand the question can anybody just give an example of $A$ and $B$ set and whats asked here. can somebody check my answer it is in the comments. should the answer be $n\choose k$×$k\choose r$ where summation is over $r$ from $1$ to $k$ and $k$ is fixed from $1$ to $n$

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    $A$ is contained in $S$ and $B$ is contained in $A$ (this means [subsets](https://en.wikipedia.org/wiki/Subset)). And by "containment is proper", the question is implying [proper subsets](http://mathworld.wolfram.com/ProperSubset.html). An example would be $A =$ {$1,2,3$} and $B=${$1,2$} for $n \ne 3$ (otherwise $A=S$) – 2017-01-19
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    can you check my comment – 2017-01-19

2 Answers 2

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Identify an ordered pair of sets $\langle B,A\rangle$ satisfying $B\subseteq A\subseteq S$ with a function $f:S\to \{0,1,2\}$.

This by stating that $B:=f^{-1}(\{0\})$ and $A:=f^{-1}(\{0,1\})$.

We are dealing with a one-to-one correspondence between these ordered pairs and these functions.

Let $W$ denote the set of functions $S\to\{0,1,2\}$ and define $U,V\subset W$ by:

  • $U:=\{f\in W\mid0\notin\text{im} f\}$
  • $V:=\{f\in W\mid1\notin\text{im} f\}$

Then $U$ corresponds with $\{\langle\varnothing,B\rangle\mid B\subseteq S\}$ and $V$ corresponds with $\{\langle B,B\rangle\mid B\subseteq S\}$.

It is obvious that $|W|=3^n$ and $|U|=|V|=2^n$ and $|U\cap V|=1$.

Observe that $\{\langle B,A\rangle\mid\varnothing\neq B\subseteq A\subseteq S\}$ corresponds with $U^c=\{f\in W\mid0\in\text{im} f\}$.

Observe that $\{\langle B,A\rangle\mid\varnothing\neq B\subsetneq A\subseteq S\}$ corresponds with $U^c\cap V^c=\{f\in W\mid0,1\in\text{im} f\}$.

It remains to find the cardinality of these sets and for the first we find :$$|U^c|=|W|-|U|=3^n-2^n$$

This agrees with your answer in the comment on the answer of David.

Can you find the cardinality of $|U^c\cap V^c|$ yourself? If not then you can take a look here:

$$|U^c\cap V^c|=|W|-|U\cup V|=|W|-|U|-|V|+|U\cap V|=3^n-2^{n+1}+1$$

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    yes drhab i was trying to make this point but i guess i wasn't doing it properly.But thanx anyways both of you. – 2017-01-23
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suppose $A$ has $k$ elements ($0 \le k \le n$).

there are $\binom n k$ possible choices for such an $A$.

To get a $B$ which contains this $A$ we require $B = A \cup C$ where $C$ is any of the $2^{n-k}$ subsets of the complement of $A$.

thus you require the sum of $\binom n k 2^{n-k}$ over the permissible values of $k$.

hint: $\binom n k = \binom n {n-k}$

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    $k\choose r$×$n\choose k$ with the summation on $r$ from $1$ to $k$ and$ k\ge1$.should this be the answer when the B is not empty. my reason is if i choose k element subset then that k cannot be 0 because that would mean an empty set and it cannot contain a nonempty set.Hence k is greater than .now from this k element if we choose any subset we are done and this can be done by choosing r element subset of k elements where r runs from 1 to k .and then we sum it over all k – 2017-01-19
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    can you check David – 2017-01-19
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    looking at subsets of size k, for each the of $\binom n k$ subsets $A$ there are $2^{n-k}$ subsets $B$ so you need $\sum_{k=0}^n \binom n k 2^{n-k}$ which evaluates to $3^n$. – 2017-01-20
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    david i am sorry i didn't get it ...can you tell what am i calculating above..is my logic wrong and in you answee does that rule out the case when B is empty – 2017-01-20
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    if $A$ is the empty set then there are $2^n$ subsets that contain it. if $A$ has one element there are $2^{n-1}$ and so on. you need to know that a set with $m$ elements has $2^m$ subsets. also there are $\binom n k$ ways of choosing a subset with $k$ elements from a set with $n$ elements. also the binomial expansion $3^n = (1+2)^n = \sum_{k=0}^n \binom n k 2^k$ – 2017-01-20
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    i am sorry i got all of that things i need to know but not the answer.but still thnx – 2017-01-20
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    i thought the answer should be$ 3^n-2^n$ – 2017-01-20
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    that is the answer to when the inclusion of $A$ in $B$ is proper since this reduces the sum to: $\sum_{k=0}^n \binom n k (2^k - 1)$ – 2017-01-20
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    In your comments you switch $A$ and $B$. In the question we have $B\subseteq A$. Concerning your last comment: $3^n-2^n$ is also the answer if the inclusion is not necessarily proper but instead $B$ (the smaller one) is demanded to be non-empty. That is more in line with the question. – 2017-01-22
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    sorry! short memory problem. i scrolled up to OP's question to get the notation consistent, but was unfortunately seduced by alphabetical order. can't edit the comment now, but thanks for pointing out the error and elaborating. – 2017-01-22
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    I make lots of these mistakes myself :-). Also $A\subseteq B$ looks more natural. – 2017-01-22