If $A, B, C, D$ are numbers in increasing order and $D, B, E$ are numbers in decreasing order, then which one of the following sequences need neither be in a decreasing nor in an increasing order?

### Concept & Logic This is an inequalities and relative ordering problem. We must establish a master chain of inequalities by combining the given sequences. ### Step-by-Step Solution * **Given:** 1) $A, B, C, D$ are in increasing order: $A < B < C < D$ 2) $D, B, E$ are in decreasing order: $D > B > E$ (which rewrites to $E < B < D$) * **Deduction:** Combine the knowns relative to $B$: * $A < B$ * $E < B$ * $B < C < D$ * Let's evaluate the options to find the unordered sequence: * (a) $E, C, D$: Since $E < B$ and $B < C < D$, we know $E < C < D$. This is strictly increasing. * (b) $E, B, C$: Since $E < B$ and $B < C$, we know $E < B < C$. This is strictly increasing. * (c) $D, B, A$: Since $A < B < D$, reversing it gives $D > B > A$. This is strictly decreasing. * (d) $A, E, C$: We know $A < B$ and $E < B$. However, no relationship is given between $A$ and $E$. One could be larger than the other. Thus, the order between $A$ and $E$ is undefined. ### Exam Strategy & Shortcut **Variable Assignment:** Assign simple numbers to fit the rules. Let $B = 10, C = 20, D = 30$. For $A < B$, let $A = 5$. For $E < B$, let $E = 8$. Now check (d): $A, E, C \implies 5, 8, 20$ (increasing). Now change $E$ to $2$ (since $2 < 10$). Check (d): $A, E, C \implies 5, 2, 20$ (neither strictly increasing nor decreasing). Since it breaks, this is the answer. ### Common Pitfall Assuming that because $A$ is the "first" variable mentioned, it must be the absolute smallest. The problem only states $A < B$; $E$ could easily be smaller than $A$. ### Final Answer Therefore, the correct answer is **$A, E, C$**.