Let $A$ be a $3 \times 3$ matrix which contains five 'a' & four 'b' then number of symmetric matrices possible is $k$, number of zeros at the end of $k!$ is _____.
Step-by-Step Solution
Key Concept: A symmetric matrix has $a_{ij} = a_{ji}$; count distinct structures where diagonal positions are constrained.
For a symmetric $3 \times 3$ matrix $\begin{pmatrix} a & d & e \\ d & b & f \\ e & f & c \end{pmatrix}$, the diagonal elements are $a$, $b$, $c$. There are three forms with diagonal '1': $\begin{pmatrix} a & a & b \\ a & a & b \\ b & b & a \end{pmatrix}$, $\begin{pmatrix} a & b & a \\ b & a & b \\ a & b & a \end{pmatrix}$, and $\begin{pmatrix} a & b & b \\ b & a & a \\ b & a & a \end{pmatrix}$, giving 3 symmetric matrices.
Correct Answer: 2