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. $e_1+e_2+O_1+O_2=5 p_3+5^3$
• 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$