A) \[n(n-1)\]
B) \[n{{(n-1)}^{2}}\]
C) \[{{n}^{2}}(n-1)\]
D) \[{{n}^{2}}{{(n-1)}^{2}}\]
Correct Answer: C
Solution :
[c] No. of triangles \[{{=}^{2n}}{{C}_{3}}{{-}^{n}}{{C}_{3}}{{-}^{n}}{{C}_{3}}\] \[=\frac{2n(2n-1)(2n-2)}{6}-\frac{2n(n-1)(n-2)}{6}\] \[=\frac{1}{3}n(n-1)(3n)={{n}^{2}}(n-1).\]You need to login to perform this action.
You will be redirected in
3 sec