# 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?

• Close, but you divided by $3!$ when you didn't need to. Note that $\binom{15}{3}=\frac{15\cdot 14\cdot 13}{3!}$. The division by $3!$ that you think you are remembering is necessary is already a part of the $\binom{15}{3}$ and didn't need to be reapplied again. Commented Nov 17, 2019 at 13:55

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