1. Permutation and Combination

Factorial For $n$ different objects, if we have a queue of all $n$ objects, then the number of queue is
$n! = n \times (n - 1) \times (n - 2) \times \dots \times 2 \times 1$
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
$P_{r}^{n} = \frac{n!}{(n - r)!}$
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
$C_{r}^{n} = \frac{P_{r}^{n}}{r!} = \frac{n!}{r! (n - r)!}$

Evaluate

C_{3}^{5} + C_{2}^{10}

by using the formula.

$\begin{array}{cl} C_{3}^{5} + C_{2}^{10} \\ = & \frac{5!}{3! (5 - 3)!} + \frac{10!}{2! (10 - 2)!} \\ = & \frac{5!}{3! 2!} + \frac{10!}{2! 8!} \\ = & \frac{5 \times 4}{2} + \frac{10 \times 9}{2} \\ = & 55 \end{array}$

Simplify

C_{3}^{n + 2} - C_{3}^{n}

$\begin{array}{cl} C_{3}^{n + 2} - C_{3}^{n} \\ = & \frac{(n + 2)!}{3! (n + 2 - 3)!} - \frac{n!}{3! (n - 3)!} \\ = & \frac{(n + 2)!}{3! (n - 1)!} - \frac{n!}{3! (n - 3)!} \\ = & \frac{n (n + 1) (n + 2)}{3!} - \frac{n (n - 1) (n - 2)}{3!} \\ = & \frac{n}{3!} [(n + 1) (n + 2) - (n - 1) (n - 2)] \\ = & \frac{n}{3!} (n^{2} + 3 n + 2 - n^{2} + 3 n - 2) \\ = & \frac{n}{3!} (6 n) \\ = & n^{2} \end{array}$

Solve

C_{n - 1}^{n} \times C_{n - 2}^{n - 1} = 12

$\begin{array}{rcl} C_{n - 1}^{n} \times C_{n - 2}^{n - 1} & = & 12 \\ \frac{n!}{(n - 1)! [n - (n - 1)]!} \times \frac{(n - 1)!}{(n - 2)! [n - 1 - (n - 2)]!} & = & 12 \\ \frac{n!}{(n - 1)! 1!} \times \frac{(n - 1)!}{(n - 2)! 1!} & = & 12 \\ n (n - 1) - 12 & = & 0 \\ n^{2} - n - 12 & = & 0 \\ (n - 4) (n + 3) & = & 0 \end{array}$

$∴ n = 4$ or $n = - 3$ . (rejected)

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