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Question #40572

Difficulty: 8/10 | 🟢 Free

The Question

Let $x_i$ be the number of ways to arrange $i+1$ identical items in $2i+3$ spaces in a row such that no two items are adjacent, for $i = 1, 2, 3$ . Let $y_j$ be the number of ways to arrange $j+1$ identical items in $2j+2$ spaces in a row such that no two items are adjacent, for $j = 1, 2, 3$ . If $M$ is a $3 \times 3$ matrix whose entries are given by $M_{i,j} = \binom{x_i + y_j}{2}$ , find the value of $\det(M)$ .

Options

A.
$-90$
B.
$90$
C.
$-120$
D.
$120$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Bridge

Practice problems that bridge the gap between foundational theory and advanced practical application.

Concepts Behind This Question

Gap Method (Non-adjacency)|Vandermonde's Identity|Determinant Product Rule

Why This Question Is Tricky

Common Mistakes

This question fuses the Gap Method with the Determinant Product Rule via Vandermonde's Identity.

The core trap lies in the parity of permutations within determinant properties.

Students often recognize that $M_{i,j}$ can be split into a product of two matrices $A$ and $B^T$ using combinatorial identities.

However, because the index in $\binom{y_j}{3-k}$ descends, the resulting Vandermonde matrix $B$ has reversed columns.

Students accustomed to combinatorial matrices yielding strictly non-negative determinants (due to the non-negative nature of arrangements) frequently overlook the $(-1)^{\lfloor n/2 \rfloor}$ sign-flip introduced by this column reversal, leading them to incorrectly choose the positive counterpart of the determinant.

Quick Facts About This Question

Question Information

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