2. The symbol $n$ ! represents the product of the positive integers from 1 to $n$. That is, $n!=n \times(n-1) \times(n-2) \times \cdots \times 3 \times 2 \times 1$. (The symbol $n$ ! is read " $n$ factorial".) For example, the value of 4 ! is 24 because $4 \times 3 \times 2 \times 1=24$.
(a) Determine the value of $\frac{7!}{5!}$.
(b) Determine the positive integer $n$ for which $98!\times 9900=n$ ! .
(c) Determine the positive integer $m$ for which $\frac{(m+2)!}{m!}=40200$.
(d) Suppose that $q$ is a positive integer and that $r$ is the number for which $(q+2)!-(q+1)!=(q!) \times r$. Show that, for every positive integer $q$, the number $r$ is an integer which is a perfect square.
2. (a) Since 7 ! $=7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1=5040$ and 5 ! $=5 \times 4 \times 3 \times 2 \times 1=120$, then $\frac{7!}{5!}=\frac{5040}{120}=42$.
Alternatively, we could note that
$$
\frac{7!}{5!}=\frac{7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{5 \times 4 \times 3 \times 2 \times 1}=\frac{7 \times 6 \times(5 \times 4 \times 3 \times 2 \times 1)}{(5 \times 4 \times 3 \times 2 \times 1)}=7 \times 6=42
$$
(b) Since $98!=98 \times 97 \times 96 \times \cdots \times 3 \times 2 \times 1$ and $9900=100 \times 99$, then
$$
98!\times 9900=(98 \times 97 \times 96 \times \cdots \times 3 \times 2 \times 1) \times(100 \times 99)=100 \times 99 \times 98 \times 97 \times 96 \times \cdots \times 3 \times 2 \times 1
$$

By definition, this is 100 ! and so $n=100$.
(c) Since $(m+2)$ ! is the product of all positive integers from 1 to $m+2$, then
$$
(m+2)!=(m+2) \times(m+1) \times m \times(m-1) \times \cdots \times 3 \times 2 \times 1=(m+2) \times(m+1) \times m!
$$

Therefore, $\frac{(m+2)!}{m!}=\frac{(m+2) \times(m+1) \times m!}{m!}=(m+2) \times(m+1)$.
Since $\frac{(m+2)!}{m!}=40200$, then $(m+2)(m+1)=40200$.
Since $40200=100 \times 402=100 \times 2 \times 201=200 \times 201$, then when $m+1=200$, we have $(m+2)(m+1)=201 \times 200=40200$.
Therefore, $m+1=200$ and so $m=199$.
(d) Since $(q+2)!-(q+1)!=(q!) \times r$, then
$$
r=\frac{(q+2)!-(q+1)!}{q!}=\frac{(q+2)!}{q!}-\frac{(q+1)!}{q!}
$$

Modelling our approach in (c), $\frac{(q+2)!}{q!}=(q+2)(q+1)$ and $\frac{(q+1)!}{q!}=\frac{(q+1) \times q!}{q!}=q+1$. Thus, $r=(q+2)(q+1)-(q+1)$.
Since $q$ is an integer, then $r$ is an integer.
Furthermore, $r=(q+1)[(q+2)-1]=(q+1)[q+1]=(q+1)^{2}$, which is a perfect square, as required.