In the diagram, each of $p, q, r, s, t, u, v$ is to be replaced with 1,2 or 3 so that $p, q$ and $r$ are all different, $q, s$ and $t$ are all different, and $r, u$ and $v$ are all different. What is the maximum possible value of $s+t+u+v$ ?
(A) 8
(B) 9
(C) 11
(D) 7
(E) 10

![](https://cdn.mathpix.com/cropped/2024_04_20_e85ef69091c57cb633f3g-034.jpg?height=366&width=314&top_left_y=1454&top_left_x=1339)
Since each of $s, t, u, v$ is equal to one of 1,2 or 3 , and $s$ and $t$ are different and $u$ and $v$ are different, then their sum cannot be any larger than $2+3+2+3=10$.

This can only happen if $s$ and $t$ are 2 and 3 in some order, and if $u$ and $v$ are 2 and 3 in some order.

![](https://cdn.mathpix.com/cropped/2024_04_20_688a3b42f69739f2e244g-070.jpg?height=377&width=314&top_left_y=1156&top_left_x=968)

But $q, s, t$ are 1,2,3 in some order and $r, u, v$ are 1,2,3 in some order.

So if $s$ and $t$ are 2 and 3 , then $q=1$. Similarly, if $u$ and $v$ are 2 and 3 , then $r=1$.

But $p, q, r$ are $1,2,3$ in some order, so we cannot have $q=r=1$.

Therefore, we cannot have $s+t+u+v=10$.

The next largest possible value of $s+t+u+v$ would be 9 .

We can construct the diagram with this value of $s+t+u+v$ by letting $s=1, t=3, u=2$, $v=3$, and proceeding as follows:

![](https://cdn.mathpix.com/cropped/2024_04_20_688a3b42f69739f2e244g-070.jpg?height=380&width=314&top_left_y=1973&top_left_x=968)

Therefore, the maximum possible value of $s+t+u+v$ is 9 .