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

Solution

Maths sir · 4 steps

Count matrices with all different entries, then remove the few where ad = bc makes the determinant zero.

1. Count the outcomes. Write the matrix as (a, b; c, d), one die per entry. There are 6⁴ = 1296 outcomes, all equally likely. All four entries different can happen in 6 × 5 × 4 × 3 = 360 ways.
2. When it is singular. The matrix is singular when ad - bc = 0, that is ad = bc. With four different numbers from 1 to 6, we need one product made in two ways from two disjoint pairs.
3. Find such products. Only two products work: 6 = 1 × 6 = 2 × 3 and 12 = 2 × 6 = 3 × 4. For each one, choose which pair sits on the main diagonal (2 ways), then order each pair (2 ways each). That gives 2 × 2 × 2 = 8 matrices per product, so 16 singular matrices in all.

   P = (360 - 16)/1296 = 344/1296 = 43/162

4. So. The probability is 43/162, which is option (d).

The idea to keep. For a singular 2 × 2 matrix with distinct entries from 1 to 6, list the products that factor in two different ways. Here only 6 and 12 do. Count the arrangements of each one carefully.

Answer from the NTA final answer key. Solution by PrepWiser.