Question

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 three digit odd numbers in which no digit is repeated and $O_2$ represent the number of four digit odd numbers in which no digit is repeated. Then

• A. $e_1=O_1,e_2=O_2$
• B. $e_1+e_2+O_1+O_2=5p_3+5^3$
• C. $\frac{e_1+e_2}{2}=\frac{O_1+O_2}{3}=6^2$
• D. $\frac{e_1+e_2}{O_1+O_2}=\frac{3}{2}$

Answer 1

Answer: (C)

$e_1=\frac{6\text{ choice }}{4\text{ choice }}\frac{6\text{ or }8}{2\text{ choice }}$
$e_1=3\times 4\times 2=24$
$e_2=\frac{2}{2}\times \frac{3}{3}\times \frac{6\text{ or }8}{2}$
$e_2=48$
$O_1=\frac{3\text{ choice }}{4\text{ choice }}\frac{1/3/5}{3\text{ choice }}$
$=3\times 4\times 3=36$
$O_2=\frac{2\text{ choice }3\text{ choice }}{4\text{ choice }}\frac{1/3/5}{3\text{ choice }}$
$=2\times 3\times 3\times 4=72$
$\frac{e_1+e_2}{2}=\frac{24+48}{2}=36\text{ and }\frac{O_1+O_2}{3}=\frac{36+72}{3}=36$