Please give hint: how to solve question below? This is a question from the IIT-JEE, an engineering entrance exam for admission into the Indian Institute of Technology.

Let $\mathscr A$ be the set of all $3\times3$ symmetric matrices all of whose entries are either 0 or 1. Five of those entries are 1 and four of them are zero.


*The number of matrices in $\mathscr A$ is
(A) 12
(B) 6
(C) 9
(D) 3


 A: So what do we need? We want the matrices to be symmetric, and contain five ones and four zeros. 
There are two cases:
1) The diagonal consists of $1,0,0$ in some order, which can be decided in $3$ ways. The rest of the elements are four ones and two zeros. Now, these can be permuted amongst themselves in three ways, since deciding the top three decides the bottom three, and the top three elements consists of two ones and one zero. Hence, the answer here is $3 \times 3=9$. 
2) The diagonal consists of $1,1,1$ (which can't be permuted). The rest of the elements are four zeros and two ones. Now, these can be permuted amongst themselves in three ways, since deciding the top three decides the bottom three, and the top three elements consists of two zeros and one one. Hence, the answer here is $3 \times 1=3$.  
In total, we have $9+3=12$ such matrices. Since you are doing the IITJEE, which is a very time tight exam, you can do the first step and directly write $12$ as the answer, since it has to be bigger than $9$ by the first step.
