how many $4\times 4$ matrices with specific entries Suppose we construct $4\times 4$ matrix with the following requirements:
Entry $1,1$ is $a$. We have to insert three elements $b$, one element $c$ and one element $d$. Other entries are $0$. How many such matrices are there?
Since there are three $b$ elements, out of $4\cdot 4 -1$ ($a$ takes up the first entry) we can choose $3$ random entries, but we do not distinguish between $b$ elements so the choices are $$\frac{{15 \choose 3}}{3!}.$$ Then we proceed to choose any of the $12$ entries for element $c$ and then $11$ choices for element $d$. Thus there are $$\frac{{15 \choose 3}}{3!}\cdot 12\cdot 11$$ such matrices. Is this correct?

You do not need to divide by $3!$, the binomial counts unordered groups. $${15 \choose 3}\cdot 12\cdot 11$$
