Permutations And Combinations Question 68
Question: There are $ m $ points on a straight line $ AB $ and $ n $ points on another line $ AC $ , none of them being the point $ A $ . Triangles are formed from these points as vertices when (i) $ A $ is excluded (ii) $ A $ is included. Then the ratio of the number of triangles in the two cases is
Options:
A) $ \frac{m+n-2}{m+n} $
B) $ \frac{m+n-2}{2} $
C) $ \frac{m+n-2}{m+n+2} $
D) None of these
Correct Answer: A
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Answer:
Solution:
$ \therefore $ Number of triangles $ =mn+\frac{1}{2}mn(m+n-2) $ $ =\frac{1}{2}mn(m+n) $ ?..(ii)
$ \therefore $ Required ratio $ =\frac{(m+n-2)}{(m+n)} $ .
BETA
