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: For a 3×3 symmetric matrix with 6 independent positions (3 diagonal + 3 upper triangular), distribute 5 'a's and 4 'b's using multinomial coefficients: C(6,5)=6, so k=6. Find trailing zeros in 6! by counting factors of 5 in 720=2⁴×3²×5¹, yielding exactly 1 factor of 5.
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