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If you only know that your poset is graded, I don't see any easy bounds, unless you have some auxiliary information. In general, the $y$'s could be all elements of rank $k$ (except $x$ itself), for example if your poset looks like this, and you fix $m=1$, $k=3$, $x=11$. Every distinct pair of elements at rank $k=3$ meet at rank $m=1$, namely, at the element "1".

At the other extreme, if you know your poset is the Boolean lattice on a ground set of $n$ elements, the answer is simple. Seeing that $x$ and $y$ are sets of $k$ elements, their meet has rank $m$ iff $|x \cap y|=m$. The number of such sets is $$ \binom{k}{m} \binom{n-k}{k-m} $$ because the $y$'s are those $k$-element sets that have $m$ elements from $x$, and $k-m$ elements from outside $x$.

However, it sounds like you have some case in between. It may be easier to answer if you can specify your case in more detail.

If you only know that your poset is graded, I don't see any easy bounds, unless you have some auxiliary information. In general, the $y$'s could be all elements of rank $k$ (except $x$ itself), for example if your poset looks like this, and you fix $m=1$, $k=3$, $x=11$. Every distinct pair of elements at rank $k=3$ meet at rank $m=1$, namely, at the element "1".

At the other extreme, if you know your poset is the Boolean lattice on a ground set of $n$ elements, the answer is simple. Seeing that $x$ and $y$ are sets of $k$ elements, their meet has rank $m$ iff $|x \cap y|=m$. Given $x$, the number of such $y$'s is $$ \binom{k}{m} \binom{n-k}{k-m} $$ because they are exactly those $k$-element sets that have $m$ elements from $x$, and $k-m$ elements from outside $x$.

However, it sounds like you have some case in between. It may be easier to answer if you can specify your case in more detail.

If you only know that your poset is graded, I don't see any easy bounds, unless you have some auxiliary information. In general, $y$ could be all elements of rank $k$ (except $x$ itself), for example if your poset looks like this, and you fix $m=1$, $k=3$, $x=11$. Every distinct pair of elements at rank $k=3$ meet at rank $m=1$, namely, at the element "1".

At the other extreme, if you know your poset is the Boolean lattice on a ground set of $n$ elements, the answer is simple. Seeing that $x$ and $y$ are sets of $k$ elements, their meet has rank $m$ iff $|x \cap y|=m$. The number of such sets is $$ \binom{k}{m} \binom{n-k}{k-m} $$ because the $y$'s are those $k$-element sets that have $m$ elements from $x$, and $k-m$ elements from outside $x$.

However, it sounds like you have some case in between. It may be easier to answer if you can specify your case in more detail.

If you only know that your poset is graded, I don't see any easy bounds, unless you have some auxiliary information. In general, the $y$'s could be all elements of rank $k$ (except $x$ itself), for example if your poset looks like this, and you fix $m=1$, $k=3$, $x=11$. Every distinct pair of elements at rank $k=3$ meet at rank $m=1$, namely, at the element "1".

At the other extreme, if you know your poset is the Boolean lattice on a ground set of $n$ elements, the answer is simple. Seeing that $x$ and $y$ are sets of $k$ elements, their meet has rank $m$ iff $|x \cap y|=m$. The number of such sets is $$ \binom{k}{m} \binom{n-k}{k-m} $$ because the $y$'s are those $k$-element sets that have $m$ elements from $x$, and $k-m$ elements from outside $x$.

However, it sounds like you have some case in between. It may be easier to answer if you can specify your case in more detail.

If you only know that your poset is graded, I don't see any easy bounds, unless you have some auxiliary information. In general, $y$ could be all elements of rank $k$, for example if your poset looks like this, and you fix $m=1$, $k=3$, $x=11$. Every pair of elements at rank $k=3$ meet at rank $m=1$, namely, at the element "1".

At the other extreme, if you know your poset is the Boolean lattice on a ground set of $n$ elements, the answer is simple. Seeing that $x$ and $y$ are sets of $k$ elements, their meet has rank $m$ iff $|x \cap y|=m$. The number of such sets is $$ \binom{k}{m} \binom{n-k}{k-m} $$ because the $y$'s are those $k$-element sets that have $m$ elements from $x$, and $k-m$ elements from outside $x$.

However, it sounds like you have some case in between. It may be easier to answer if you can specify your case in more detail.

If you only know that your poset is graded, I don't see any easy bounds, unless you have some auxiliary information. In general, $y$ could be all elements of rank $k$ (except $x$ itself), for example if your poset looks like this, and you fix $m=1$, $k=3$, $x=11$. Every distinct pair of elements at rank $k=3$ meet at rank $m=1$, namely, at the element "1".

At the other extreme, if you know your poset is the Boolean lattice on a ground set of $n$ elements, the answer is simple. Seeing that $x$ and $y$ are sets of $k$ elements, their meet has rank $m$ iff $|x \cap y|=m$. The number of such sets is $$ \binom{k}{m} \binom{n-k}{k-m} $$ because the $y$'s are those $k$-element sets that have $m$ elements from $x$, and $k-m$ elements from outside $x$.

However, it sounds like you have some case in between. It may be easier to answer if you can specify your case in more detail.