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Question #40587

Difficulty: 8/10 | 🟢 Free

The Question

Let $P(x) = \begin{vmatrix} x-4 & -1 & -1 & -1 \\ -1 & x-4 & -1 & -1 \\ -1 & -1 & x-4 & -1 \\ -1 & -1 & -1 & x-4 \end{vmatrix}$ . Let $n_1, n_2, n_3, n_4$ be the roots of the equation $P(x) = 0$ . A set of $N = \sum_{i=1}^4 n_i$ distinct items is to be divided into four unlabeled groups such that the sizes of the groups are $n_1, n_2, n_3,$ and $n_4$ . Let $W$ be the number of ways to perform this division. Find the value of $W$ .

Options

A.
3203200
B.
19219200
C.
800800
D.
2627625

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

Division into Groups|Multinomial Theorem|Factor Theorem in Determinants

Why This Question Is Tricky

Common Mistakes

The problem fuses Linear Algebra (Determinants) with Combinatorics (Division into Groups).

The primary trap lies in the layered dependency: students must first correctly apply row operations or the Factor Theorem to extract the roots of the characteristic-like polynomial, which act as the group sizes.

The lethal collision occurs in the combinatorics phase: because the determinant yields a repeated root ( $x=3$ with multiplicity 3), three of the groups have identical sizes.

Students frequently calculate the multinomial coefficient $\frac{16!}{7!(3!)^3}$ but forget to divide by $3!$ to account for the unlabeled nature of identical-sized groups, leading them straight to the trap answer 19219200.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Permutations and Combinations - Class 11
Concept	Division into Groups
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free