Problem 435

Let Anyu, Brad, Chi, and Diego be represented by $A$, $B$, $C$, and $D$, respectively, and so their original
order is $ABCD$.

When rearranged, $A$ is not in the
1$^{\text{st}}$ position, and so
there are exactly 3 cases to consider: $A$ is in the 2$^{\text{nd}}$ position, or $A$ is in the 3$^{\text{rd}}$ position, or $A$ is in the 4$^{\text{th}}$ position.

For each of these 3 cases, we count the number of ways to arrange $B$, $C$, and $D$.

Case 1: $A$ is in
the 2$^{\text{nd}}$ position

Since $A$ is in the 2$^{\text{nd}}$ position, $B$ can be in any of the other 3 positions
$(\$1^{\text{st}}, 3^{\text{rd}} \text{}$ or }
$4^{\text{th}}\$).$

If $B$ is in the 1$^{\text{st}}$ position, then there is
exactly one possible rearrangement: $BADC$ (since $C$ and $D$ cannot be in the 3$^{\text{rd}}$ and 4$^{\text{th}}$ positions
respectively).

If $B$ is in the 3$^{\text{rd}}$ position, then there is
exactly one possible rearrangement: $DABC$ (since $D$ cannot be in the 4$^{\text{th}}$ position).

If $B$ is in the 4$^{\text{th}}$ position, then there is
exactly one possible rearrangement: $CADB$ (since $C$ cannot be in the 3$^{\text{rd}}$ position).

Thus there are exactly 3 possible rearrangements when $A$ is in the 2$^{\text{nd}}$ position.

Case 2: $A$ is in
the 3$^{\text{rd}}$ position

Since $A$ is in the 3$^{\text{rd}}$ position, $C$ can be in any of the other 3
positions.

In a manner similar to Case 1, it can be shown that there are 3 possible
rearrangements in this case: $CDAB$,
$DCAB$, and $BDAC$.

Case 3: $A$ is in
the 4$^{\text{th}}$ position

Since $A$ is in the 4$^{\text{th}}$ position, $D$ can be in any of the other 3
positions.

Similarly, there are 3 possible rearrangements in this case: $DCBA$, $CDBA$, and $BCDA$.

So that each person is not in their original position, the four
friends can rearrange themselves in $3+3+3=9$ different ways.

(Such a rearrangement of a list in which no element appears in its
original position is called a derangement.)