Four dice are thrown simultaneously and the numbers shown on these dice are recorded in 2 × 2 matric

Q: Why do we use the complementary approach? Counting non-singular matrices directly seems more intuitive.

It is a classic trap! Counting non-singular matrices directly is a combinatorial nightmare because there are so many combinations that satisfy $ad \neq bc$. By counting the singular ones ($ad = bc$), we isolate the specific, rare cases where the determinant vanishes. It is much cleaner to subtract these few cases from the total.

Q: How do we ensure all entries are distinct when forming the matrix?

Great question. We treat this as a two-step process. First, we select 4 distinct numbers from the 6 available using combinations ($^6C_4$). Then, we arrange these 4 chosen numbers into the 4 slots of the matrix, which is a permutation ($4!$). Multiplying these gives the total number of matrices with distinct entries.

Q: Are there any other sets of 4 numbers that satisfy $ad = bc$ besides the ones we found?

Excellent attention to detail. We must be exhaustive. We need $ad = bc$ with distinct numbers. If you test combinations like $\{1, 2, 3, 4\}$, $\{1, 2, 4, 5\}$, etc., you will find that only $\{1, 2, 3, 6\}$ and $\{2, 3, 4, 6\}$ work. Any other combination fails to produce equal products. Always verify your cases systematically!

Question

Four dice are thrown simultaneously and the numbers shown on these dice are recorded in $2 	imes 2$ matrices. The probability that such formed matrices have all different entries and are non-singular, is :

Options:

(A) $45/162$

(B) $23/81$

(C) $22/81$

(D) $43/162$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineer. Today, we are not just solving a probability problem; we are stepping into the shoes of an architect designing a system of constraints.

Imagine you have four dice in your hand. You roll them, and the numbers that appear are the building blocks of a $2 	imes 2$ matrix:

$$A = \begin{pmatrix} a & b \ c & d \end{pmatrix}$$

We want to find the probability that this matrix is both 'distinct' (all entries are unique) and 'non-singular' (the determinant is non-zero).

### The Foundation of the Sample Space

Every great probability problem begins with the total sample space. When you roll four dice, each die is an independent agent of chaos, capable of showing any integer from $1$ to $6$.

Since there are four dice, the total number of outcomes is:

$$6 	imes 6 	imes 6 	imes 6 = 6^4 = 1296$$

This is our denominator. It is the vast ocean of possibilities from which we must fish out our specific, favorable outcomes.

### The Constraint of Uniqueness

Now, we impose our first constraint: the entries $a, b, c, d$ must be distinct. This is a beautiful combinatorial challenge.

We are picking a set of four unique numbers from the six available on the dice. The number of ways to choose these four numbers is given by the combination formula:

$$^6C_4 = \frac{6 	imes 5}{2 	imes 1} = 15$$

Once we have our set of four numbers, we must arrange them into the four slots of our matrix. The number of ways to arrange four distinct items is $4! = 24$.

Therefore, the total number of matrices with distinct entries is $15 	imes 24 = 360$. This is our new, restricted universe.

### The Singularity Trap

Here is where the problem gets thrilling. We need the matrix to be non-singular, meaning the determinant $|A| = ad - bc 
eq 0$.

If you try to count all the non-singular matrices directly, you will be lost in a forest of inequalities. Instead, we use the power of complementary counting.

We will find the number of singular matrices (where $ad - bc = 0$, or $ad = bc$) and subtract them from our total of $360$. This is the elegant path.

### The Hunt for Equal Products

We need to find sets of four distinct numbers from $\{1, 2, 3, 4, 5, 6\}$ such that the product of two equals the product of the other two.

**Case 1: The product is $6$.** 
We can use the set $\{1, 2, 3, 6\}$ because $1 	imes 6 = 2 	imes 3 = 6$.

The pair $(a, d)$ can be $(1, 6)$ or $(6, 1)$ (2 ways). The pair $(b, c)$ can be $(2, 3)$ or $(3, 2)$ (2 ways). Furthermore, we can swap the pairs entirely, making $(a, d)$ the $(2, 3)$ pair and $(b, c)$ the $(1, 6)$ pair (2 ways).

Thus, we have $2 	imes 2 	imes 2 = 8$ singular matrices for this case.

**Case 2: The product is $12$.** 
We can use the set $\{2, 3, 4, 6\}$ because $2 	imes 6 = 3 	imes 4 = 12$.

Using the same logic as Case 1, we find another $8$ singular matrices. If you check other combinations, you will find no other sets of four distinct numbers satisfy the condition $ad = bc$.

Thus, the total number of singular matrices is $8 + 8 = 16$.

### Final Calculation

We started with $360$ matrices that had distinct entries. We discovered that $16$ of them are singular (the 'collapse' of the matrix).

Therefore, the number of favorable matrices is $360 - 16 = 344$. The final probability is the ratio of favorable outcomes to the total sample space:

$$P(E) = \frac{344}{1296}$$

By dividing both the numerator and the denominator by $8$, we arrive at the beautiful, simplified result:

$$\mathbf{\frac{43}{162}}$$

You have navigated the constraints, avoided the traps, and arrived at the truth. This is the essence of JEE Advanced mathematics—systematic, logical, and deeply rewarding.

Animated Solution for Mathematics - Probability: Four dice are thrown simultaneously and the numbers shown on these dice are recorded in $2 \times 2$ matrices. The probability that such formed matrices have all different entries and are non-singular, is :

Select Answer:

Visualized Solution

The Physical Setup (Sample Space)

The Matrix Construction

The Constraints

Total Distinct Matrices (Selection)

Total Distinct Matrices (Arrangement)

The Complementary Approach (Singular Matrices)

Finding Equal Products (Case 1)

Counting Case 1 Matrices

Finding Equal Products (Case 2)

Total Singular Matrices

Favorable Outcomes

Final Probability

The Sigma Insight: Classical Definition of Probability

Analyzing the Setup

The Foundation of the Sample Space

The Constraint of Uniqueness

The Singularity Trap

The Hunt for Equal Products

Final Calculation

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