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

Probability PYQ — JEE Main 2021 · 22 Jul · Shift 2 · Q27 with Solution

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Correct answer: D

Total number of outcomes

Each of the 4 dice can show any of the numbers $1$ to $6$ . So total possible $2\times 2$ matrices are

$6^4 = 1296.$

Let the matrix be

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

where $a,b,c,d\in\{1,2,3,4,5,6\}$ .

We need the probability that:

all four entries are different,
the matrix is non-singular, i.e. $\det(A)\neq 0$ .

So we count matrices with all entries distinct and subtract those which are singular.

Count matrices with all different entries

We must place 4 distinct numbers from $1$ to $6$ into the 4 positions.

Number of such matrices:

${}^6P_4 = 6\cdot 5\cdot 4\cdot 3 = 360.$

Among these, count singular matrices

A matrix is singular when

$ad-bc=0 \quad\Rightarrow\quad ad=bc.$

Also, $a,b,c,d$ are all distinct.

So we need ordered quadruples $(a,b,c,d)$ of distinct numbers from $1$ to $6$ such that

$ad=bc.$

We now find all possibilities.

Possible equal products using distinct numbers from $1$ to $6$

List products of two distinct numbers from $1$ to $6$ :

\begin{aligned} 1\cdot 2&=2, &1\cdot 3&=3, &1\cdot 4&=4, &1\cdot 5&=5, &1\cdot 6&=6,\\ 2\cdot 3&=6, &2\cdot 4&=8, &2\cdot 5&=10, &2\cdot 6&=12,\\ 3\cdot 4&=12, &3\cdot 5&=15, &3\cdot 6&=18,\\ 4\cdot 5&=20, &4\cdot 6&=24, &5\cdot 6&=30. \end{aligned}

For $ad=bc$ with all four entries distinct, we need two different pairs of distinct numbers having the same product and no common element.

From the list, the only repeated products are:

$6 = 1\cdot 6 = 2\cdot 3$
$12 = 2\cdot 6 = 3\cdot 4$

But for product $12$ , the pairs $(2,6)$ and $(3,4)$ are disjoint, so this works. For product $6$ , the pairs $(1,6)$ and $(2,3)$ are also disjoint, so this works.

Thus the only sets of four distinct numbers that can make a singular matrix are:

$\{1,2,3,6\}$ with pairing $(1,6)$ and $(2,3)$ ,
$\{2,3,4,6\}$ with pairing $(2,6)$ and $(3,4)$ .

Count ordered matrices from each product pattern

For a singular matrix, we need

$ad = bc.$

Take the first case: pairs $(1,6)$ and $(2,3)$ .

Now:

$(a,d)$ can be $(1,6)$ or $(6,1)$ : $2$ ways,
$(b,c)$ can be $(2,3)$ or $(3,2)$ : $2$ ways.

So this gives

$2\times 2 = 4$ matrices.

Also we may swap which factor pair goes to $(a,d)$ and which to $(b,c)$ :

$(a,d)$ from $(2,3)$ and $(b,c)$ from $(1,6)$ ,

again giving $4$ matrices.

Hence total from product $6$ :

$8.$

Similarly, from product $12$ using pairs $(2,6)$ and $(3,4)$ , total matrices:

$8.$

Therefore total singular matrices with all entries distinct:

$8+8=16.$

Count favorable matrices

Matrices with all entries distinct and non-singular:

$360-16=344.$

Hence required probability is

$\frac{344}{1296} = \frac{43}{162}.$

Compare with options

$\frac{43}{162}$ corresponds to Option D.

Comparison with stored answer

Stored correct answer is D, which matches our result.

More from Probability

View written solutionFree

Correct answer: D

Total number of outcomes

Each of the 4 dice can show any of the numbers $1$ to $6$ . So total possible $2\times 2$ matrices are

$6^4 = 1296.$

Let the matrix be

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

where $a,b,c,d\in\{1,2,3,4,5,6\}$ .

We need the probability that:

all four entries are different,
the matrix is non-singular, i.e. $\det(A)\neq 0$ .

So we count matrices with all entries distinct and subtract those which are singular.

Count matrices with all different entries

We must place 4 distinct numbers from $1$ to $6$ into the 4 positions.

Number of such matrices:

${}^6P_4 = 6\cdot 5\cdot 4\cdot 3 = 360.$

Among these, count singular matrices

A matrix is singular when

$ad-bc=0 \quad\Rightarrow\quad ad=bc.$

Also, $a,b,c,d$ are all distinct.

So we need ordered quadruples $(a,b,c,d)$ of distinct numbers from $1$ to $6$ such that

$ad=bc.$

We now find all possibilities.

Possible equal products using distinct numbers from $1$ to $6$

List products of two distinct numbers from $1$ to $6$ :

\begin{aligned} 1\cdot 2&=2, &1\cdot 3&=3, &1\cdot 4&=4, &1\cdot 5&=5, &1\cdot 6&=6,\\ 2\cdot 3&=6, &2\cdot 4&=8, &2\cdot 5&=10, &2\cdot 6&=12,\\ 3\cdot 4&=12, &3\cdot 5&=15, &3\cdot 6&=18,\\ 4\cdot 5&=20, &4\cdot 6&=24, &5\cdot 6&=30. \end{aligned}

For $ad=bc$ with all four entries distinct, we need two different pairs of distinct numbers having the same product and no common element.

From the list, the only repeated products are:

$6 = 1\cdot 6 = 2\cdot 3$
$12 = 2\cdot 6 = 3\cdot 4$

But for product $12$ , the pairs $(2,6)$ and $(3,4)$ are disjoint, so this works. For product $6$ , the pairs $(1,6)$ and $(2,3)$ are also disjoint, so this works.

Thus the only sets of four distinct numbers that can make a singular matrix are:

$\{1,2,3,6\}$ with pairing $(1,6)$ and $(2,3)$ ,
$\{2,3,4,6\}$ with pairing $(2,6)$ and $(3,4)$ .

Count ordered matrices from each product pattern

For a singular matrix, we need

$ad = bc.$

Take the first case: pairs $(1,6)$ and $(2,3)$ .

Now:

$(a,d)$ can be $(1,6)$ or $(6,1)$ : $2$ ways,
$(b,c)$ can be $(2,3)$ or $(3,2)$ : $2$ ways.

So this gives

$2\times 2 = 4$ matrices.

Also we may swap which factor pair goes to $(a,d)$ and which to $(b,c)$ :

$(a,d)$ from $(2,3)$ and $(b,c)$ from $(1,6)$ ,

again giving $4$ matrices.

Hence total from product $6$ :

$8.$

Similarly, from product $12$ using pairs $(2,6)$ and $(3,4)$ , total matrices:

$8.$

Therefore total singular matrices with all entries distinct:

$8+8=16.$

Count favorable matrices

Matrices with all entries distinct and non-singular:

$360-16=344.$

Hence required probability is

$\frac{344}{1296} = \frac{43}{162}.$

Compare with options

$\frac{43}{162}$ corresponds to Option D.

Comparison with stored answer

Stored correct answer is D, which matches our result.

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