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Q.

There are ‘m’ points on a straight line AB and 'n' points on another straight line AC in which A is not in included. By joining these points triangles are constructed. case i) When A is included. case ii) When A is not included, the ratio of number of triangles in both cases is

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a

$\frac{m + n - 2}{2}$

b

$\frac{m + n - 2}{m + n + 2}$

c

$\frac{m + n - 2}{m + n}$

d

$\frac{m + n + 2}{m + n - 2}$

answer is A.

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Detailed Solution

When A is not included, the number of triangles formed= $^{m} C_{2} .^{n} C_{1} +^{m} C_{1} .^{n} C_{2}$ When A is included, the number of triangles formed = $^{m} C_{2} .^{n} C_{1} +^{m} C_{1} .^{n} C_{2} +^{m} C_{1} .^{n} C_{1}$

Ratio= $\frac{^{m} C_{2} .^{n} C_{1} +^{m} C_{1} .^{n} C_{2}}{^{m} C_{2} .^{n} C_{1} +^{m} C_{1} .^{n} C_{2} +^{m} C_{1} .^{n} C_{1}} = \frac{\frac{m n}{2} (m - 1 + n - 1)}{\frac{m n}{2} (m - 1 + n - 1 + 2)} = \frac{m + n - 2}{m + n}$

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There are ‘m’ points on a straight line AB and 'n' points on another straight line AC in which A is not in included. By joining these points triangles are constructed. case i) When A is included. case ii) When A is not included, the ratio of number of triangles in both cases is