Suppose that the five distinct numbers that Joshua chooses are $V, W, X, Y, Z$, and that $V<W<X<Y<Z$. We want to assign these to $p, q, r, s, t$ so that $p<s$ and $q<s$ and $r<t$ and $s<t$. First, we note that $t$ must be the largest of $p, q, r, s, t$. This is because $r<t$ and $s<t$, and because $p<s$ and $q<s$, we get $p<s<t$ and $q<s<t$, so $p<t$ and $q<t$. Since $t$ is the largest, then $Z$ must be $t$. Now neither $p$ nor $q$ can be the second largest of the numbers (which is $Y$), since $p$ and $q$ are both smaller than $s$ and $t$. Therefore, there are two cases: $Y=r$ or $Y=s$. Case 1: $Y=r$ We have $Y=r$ and $Z=t$. This leaves $V, W, X$ (which satisfy $V<W<X$) to be assigned to $p, q$, s (which satisfy $p<s$ and $q<s$). Since $X$ is the largest of $V, W, X$ and $s$ is the largest of $p, q, s$, then $X=s$. This leaves $V, W$ to be assigned to $p, q$. Since there is no known relationship between $p$ and $q$, then there are 2 possibilities: either $V=p$ and $W=q$, or $V=q$ and $W=p$. Therefore, if $Y=r$, there are 2 possible ways to assign the numbers. Case 2: $Y=s$ We have $Y=s$ and $Z=t$. This leaves $V, W, X$ (which satisfy $V<W<X$) to be assigned to $p, q, r$. There is no known relationship between $p, q, r$. Therefore, there are 3 ways to assign one of $V, W, X$ to $p$. For each of these 3 ways, there are 2 ways of assigning one of the two remaining numbers to $q$. For each of these $3 \times 2$ ways, there is only 1 choice for the number assigned to $r$. Overall, this gives $3 \times 2 \times 1=6$ ways to do this assignment. Therefore, if $Y=s$, there are 6 possible ways to assign the numbers. Having examined the two possibilities, there are $2+6=8$ different ways to assign the numbers.