Three and four digit numbers are formed from the digits 1,3,5,6,8. if $e_{1}$ is number of three digit even numbers with no digit repeated and $e_{2}$ is number of four digit even numbers with no digit repeated. Also $O_{1}$ represent the number of four digit odd numbers in which no digit is repeated. Then

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$e_{1} = O_{1}, e_{2} = O_{2}$

$e_{1} + e_{2} + O_{1} + O_{2} = 5 p_{3} + 5^{3}$

$\frac{e_{1} + e_{2}}{2} = \frac{O_{1} + O_{2}}{3} = 6^{2}$

$\frac{e_{1} + e_{2}}{O_{1} + O_{2}} = \frac{3}{2}$

answer is C.

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Detailed Solution

$e_{1} = no . of 3 digit even numbers furned using$

$e_{1} = 2 \times 4 p_{2} \Rightarrow e_{1} = 24 e_{2} = No . of 4 digit even numbers = 2 (4 p_{3}) \Rightarrow e_{2} = 48 o_{1} = No . of 4 digit odd numbers$

$∴ o_{1} = 3 (4 p_{3}) = 3 (24) = 72 o_{2} = No . of 3 digit odd numbers = 3 (4 p_{2}) = 36 e_{1} + e_{2} + o_{1} + o_{2} = 24 + 48 + 72 + 36 \frac{e_{1} + e_{2}}{2} = \frac{24 + 48}{2} = 36, \frac{o_{1} + o_{3}}{3} = \frac{72 + 36}{3} = 36$