- Factorial For $n$ different objects, if we have a queue of all $n$ objects, then the number of queue is
\begin{equation*}
n!=n\times(n-1)\times(n-2)\times\cdots\times2\times1
\end{equation*} - Permutation For $n$ different objects, if we select $r$ objects out of these $n$ objects which the order of the selected objects matters, then the number of ways is
\begin{equation*}
P^n_r = \frac{n!}{(n-r)!}
\end{equation*} - Combination For $n$ different objects, if we select $r$ objects out of these $n$ objects which the order of the selected objected does not matter, then the number of ways is
\begin{equation*}
C^n_r = \frac{P^n_r}{r!} = \frac{n!}{r!(n-r)!}
\end{equation*}
$\begin{array}{cl}
& C^5_3+C^{10}_2 \\
= & \dfrac{5!}{3!(5-3)!}+\dfrac{10!}{2!(10-2)!} \\
= & \dfrac{5!}{3!2!}+\dfrac{10!}{2!8!} \\
= & \dfrac{5\times 4}{2}+\dfrac{10\times 9}{2} \\
= & 55
\end{array}$
$\begin{array}{cl}
& C^{n+2}_3-C^n_3 \\
= & \dfrac{(n+2)!}{3!(n+2-3)!}-\dfrac{n!}{3!(n-3)!} \\
= & \dfrac{(n+2)!}{3!(n-1)!}-\dfrac{n!}{3!(n-3)!} \\
= & \dfrac{n(n+1)(n+2)}{3!}-\dfrac{n(n-1)(n-2)}{3!} \\
= & \dfrac{n}{3!}[(n+1)(n+2)-(n-1)(n-2)] \\
= & \dfrac{n}{3!}(n^2+3n+2-n^2+3n-2) \\
= & \dfrac{n}{3!}(6n) \\
= & n^2
\end{array}$
$\begin{array}{rcl}
C^n_{n-1}\times C^{n-1}_{n-2} & = & 12 \\
\dfrac{n!}{(n-1)![n-(n-1)]!} \times \dfrac{(n-1)!}{(n-2)![n-1-(n-2)]!} & = & 12 \\
\dfrac{n!}{(n-1)!1!} \times \dfrac{(n-1)!}{(n-2)!1!} & = & 12 \\
n(n-1) -12 & = & 0 \\
n^2-n-12 & = & 0 \\
(n-4)(n+3) & = & 0
\end{array}$
$\therefore n=4$ or $n=-3$. (rejected)
