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Question #41869

Difficulty: 8/10 | 🟢 Free

The Question

Let $S$ be the set of all sequences of length 2 formed by elements from $\{1, 2\}$ . Define a $2 \times 2$ matrix $P$ such that $P_{ij}$ is the number of sequences in $S$ starting with $i$ and ending with $j$ that are non-decreasing. Similarly, define a $2 \times 2$ matrix $Q$ such that $Q_{ij}$ is the number of sequences in $S$ starting with $i$ and ending with $j$ that are non-increasing. If $M = P^6 + 6P^5Q + 15P^4Q^2 + 20P^3Q^3 + 15P^2Q^4 + 6PQ^5 + Q^6$ , find the value of $\text{Tr}(M)$ .

Options

A.
608
B.
730
C.
668
D.
728

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Monster

Ultra-high difficulty questions designed to test your limit, stamina, and deep conceptual understanding.

Concepts Behind This Question

String Method (Adjacency)|Matrix Multiplication Non-Commutativity|Symmetry vs. Antisymmetry

Why This Question Is Tricky

Common Mistakes

The question tests the student's ability to translate combinatorial constraints (adjacency and sequence ordering) into matrix representations, followed by a critical test of matrix algebra properties.

The primary trap lies in the expression for $M$ , which mimics the binomial expansion of $(P+Q)^6$ .

Students who rote-apply the binomial theorem will incorrectly assume $M = (P+Q)^6$ and compute the trace using the eigenvalues of $P+Q$ (yielding $3^6 + 1^6 = 730$ ).

However, because the adjacency matrices $P$ and $Q$ do not commute ( $PQ \neq QP$ ), the binomial expansion is invalid.

The student must recognize this non-commutativity, derive the general form of $P^a Q^b$ , and compute the trace linearly to bypass the trap and find the true value.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Permutations and Combinations - Class 11
Concept	String Method (Adjacency)
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free