Consider the partial order $(\{0,1\}^n,\leq)$, where $\{0,1\}^n$ is the set of all length-$n$-strings over $0$,$1$. For $x,y\in\{0,1\}^n$ we define $x\leq y$ if $x_i\leq y_i$ for all $1\leq i\leq n$. For example, $(1,0,0)\leq (1,0,1)$ but $(1,0,0)\not\leq (0,1,0)$.
What is the largest chain of $(\{0,1\}^n,\leq)$?
On
You can also use Mirsky's Theorem. Robert Z's example shows that the poset is of height at least $n+1$. Since the poset can be written as a union of $n+1$ antichains $$A_i:=\big\{x\in \{0,1\}^n\,|\,x\text{ has }i\text{ ones}\big\}$$ for $i=0,1,2,\ldots,n$, the height of this poset must be exactly $n+1$.
Show that the length of the longest chain in $(\{0,1\}^n,\leq)$ is $n+1$. An example of longest chain is $$(0,0,0,\dots,0,0)\leq (1,0,0,\dots,0,0)\leq (1,1,0,\dots,0,0)\\ \leq (1,1,1,\dots,0,0)\leq \cdots \leq (1,1,1,\dots,1,0)\leq (1,1,1,\dots,1,1).$$ Note that if $X_1\leq X_2\leq \dots \leq X_m$ is a chain then $0\leq |X_1|< |X_2|< \dots < |X_m|\leq n$ where, for any $X\in\{0,1\}^n$, $|X|$ denotes the number of $1$s in $X$. Does this property imply that there are no chains of length greater than $n+1$?