If (^{n}P_{3} – 6(^{n}C_{4}) = 0), find the value of n.
-
A.
5 -
B.
6 -
C.
7 -
D.
8
Correct Answer: Option C
Explanation
(^{n}P_3 – 6(^{n}C_{4})=0\frac{n!}{(n-3)!}-6left( frac{n!}{(n-4)!4!}right)=0\frac{n!}{(n-3)!}=6left(frac{n!}{(n-4)!4!}right)\n!((n-4)!4!)=6n!(n-3)!\((n-4)!4!)=6(n-3)!\frac{(n-4)!}{(n-3)!}=frac{6}{4!}\frac{(n-4)!}{(n-3)(n-4)!}=frac{6}{4 times 3times 2times 1}\frac{1}{(n-3)}=]frac{1}{4}\n-3=4\n=4+3\n=7)