If (frac{^{n}C_{3}}{^{n}P_{2}} = 1), find the value of n.
-
A.
8 -
B.
7 -
C.
6 -
D.
5
Correct Answer: Option A
Explanation
(^{n}C_{3} = frac{n!}{(n – 3)! 3!})
(^{n}P_{2} = frac{n!}{(n – 2)!})
(frac{^{n}C_{3}}{^{n}P_{2}} = frac{n!}{(n – 3)! 3!} ÷ frac{n!}{(n – 2)!})
(frac{n!}{(n – 3)! 3!} times frac{(n – 2)!}{n!} = frac{(n – 2)!}{(n – 3)! 3!})
Note that ((n – 2)! = (n – 2) times (n – 2 – 1)! = (n – 2)(n – 3)!)
(frac{(n – 2)(n – 3)!}{(n – 3)! 3!} = 1)
(frac{n – 2}{3!} = 1 implies n – 2 = 6)
(n = 2 + 6 = 8)