<s>[INST] Find the number of all ordered triplets $(A,B,C)$ of subsets of $X$ such that $A$ is a subset of $B$ and $B$ is a proper subset of $C.$ Let $X$ be as set containing $n$ elements.Find the number of all ordered triplets $(A,B,C)$ of subsets of $X$ such that $A$ is a subset of $B$ and $B$ is a proper subset of $C.$

I have no idea how to solve this problem.Please help me.Thanks.
 [/INST] First let's solve this simplier problem : What's the number $\mathcal{N}$ of sets of $X$ such that $|A| = a$, $|B| = b$ and $|C| = c$ with $n \geq a>b>c >0$?
There is $n\choose a$ way to choose $A$, then $a\choose b$ ways to choose $B$ inside $A$, then $b\choose c$ way to choose $C$ inside $B$. This way, we have all the possible set without double. 
$$\mathcal{N}(a,b,c) = {n\choose a}{a \choose b}{b\choose c}$$
So now we can answer the original question (with $n \geq 3$):
$$\mathcal{N} = \sum_{a=3}^n \sum_{b=2}^a \sum_{c=1}^b {n\choose a}{a \choose b}{b\choose c}$$
$$=\sum_{a=3}^n \left( {n\choose a} \sum_{b=2}^a \left( {a \choose b} \sum_{c=1}^b {b\choose c} \right) \right)$$
But with classical binomial formula, this is also equal to 
$$=\sum_{a=3}^n \left( {n\choose a} \sum_{b=2}^a \left( {a \choose b} \left( \sum_{c=0}^b {b\choose c} - 1 \right)\right) \right)$$
$$=\sum_{a=3}^n \left( {n\choose a} \sum_{b=2}^a \left( {a \choose b} ( 2^b -1 ) \right) \right)$$
$$=\sum_{a=3}^n \left( {n\choose a} \left( \sum_{b=2}^a  {a \choose b} 2^b - \sum_{b=2}^a  {a \choose b}  \right) \right)$$
$$=\sum_{a=3}^n \left( {n\choose a} \left( \sum_{b=0}^a  {a \choose b} 2^b -( 1 + 2a ) - \sum_{b=0}^a  {a \choose b} + (1+a) \right) \right)$$
$$=\sum_{a=3}^n \left( {n\choose a} \left( 3^a-2^a-a \right) \right)$$
$$=\sum_{a=0}^n {n\choose a} 3^a - (1+3n+9\frac{n(n-1)}{2}) -\sum_{a=0}^n {n\choose a} 2^a + (1+2n+4\frac{n(n-1)}{2}) - \sum_{a=0}^n {n\choose a} a + (1+2n) $$
Finally we get
$$= 4^n -3^n - 2^{n-1}n -5\frac{n(n-1)}{2} +n+1$$
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