There are \( \mathrm{m} \) points on a straight line \( \mathrm{AB} \& \mathrm{n} \) points on t...
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There are \( \mathrm{m} \) points on a straight line \( \mathrm{AB} \& \mathrm{n} \) points on the line \( \mathrm{AC} \) none of them being the point \( \mathrm{A} \).
Triangles are formed with these points as vertices, when
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(i) \( \mathrm{A} \) is excluded
(ii) \( \mathrm{A} \) is included. The ratio of number of triangles in the two cases is:
(A) \( \frac{m+n-2}{m+n} \)
(B) \( \frac{m+n-2}{m+n-1} \)
(C) \( \frac{m+n-2}{m+n+2} \)
(D) \( \frac{\mathrm{m}(\mathrm{n}-1)}{(\mathrm{m}+1)(\mathrm{n}+1)} \)
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