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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?

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    $\begingroup$ 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. $\endgroup$
    – JMoravitz
    Commented Nov 17, 2019 at 13:55

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

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