For what of n is

^{n+1}C_{3}= 4(^{n}C_{3})?-
**A.**

6 -
**B.**

5 -
**C.**

4 -
**D.**

3

##### Correct Answer: Option D

##### Explanation

(^{n+1}C_3 = 4(^nC_3)\frac{(n+1)!}{(n+1-3)!3!} = 4left(frac{n!}{(n-3)!3!}right)\frac{(n+1)n!}{(n-2)(n-3)!}=4left(frac{n!}{n-3!}right)\=frac{n+1}{n-2}=frac{4}{1}\n+1 = 4(n-2)\n+1 = 4n-8\-3n = -9\frac{-9}{-3}\n=3)